E11 - Cấp Tốc 789+ Toán.pdf

CT1 CT1 Chu de 1 Ham s5 8 1 1 Tinh dan di~u cua ham so 8 1 2 Cl;l''''c tri ham so 13 1 3 Gia tri I6n nhat, gia tri nho nhat 19 1 4 Duemg ti~m c~n cua do thi ham so 20 1 5 Do thi ham so chua dau gia tri t[.]

CT1CT1

Chu de 1 Ham s5 8

1.1 Tinh dan di~u cua ham so 8

1.2 Cl;l'c tri ham so 13

1.3 Gia tri I6n nhat, gia tri nho nhat 19

1.4 Duemg ti~m c~n cua do thi ham so 20

1.5 Do thi ham so chua dau gia tri tuy~t doi 21

1.6 Tiep tuyen 22

1.7 Tuang giao do tht 23

Chu de 2 Liiy thira, mii va logarit 28

2.1 Ham so lily thua 28

2.2 Ham so mil 28

2.3, Logarit va ham so logarit 29

2.4 Phuang trlnh mil 30

2.5 Phuang trlnh logarit 32

2.6 Bat phuO'Ilg trlnh mil 33

2.7 Bat phuang trlnh logarit 34

2.8 Bai to an lai suat ngan hang 36

Chu de 3 Nguyen ham- tfch phan- Un.g d\lng tich phan 38

3.1 Bang nguyen ham cua m<)t so ham thuang g~p , 38

3.2 Cac phuO'Ilg phap tinh nguyen ham 38

3.3 Cac phuO'Ilg phap tinh tich phan 41

3.4 Tich phan cac ham so so cap co ban 43

3.5 Ung dvng tich phan 45

3.6 M<)t so phuang phap tinh tich phan ham an 50

3.7 M<)t so thu thu~t su dvng casio 51

Chu de 4 Sd phuc 53

4.1 Kien thuc co ban 53

4.2 Cac phep to an co ban so phuc 53

4 3 T c;tp Q'P 1em 1eu 1en sop uc A h d' A? b' A? d' J:: AI h r 55 4.4 Phuang trlnh b~c hai v6i h~ so thl;l'c 59

4.5 M<)t so bai toan lien quan den min- max so phuc 60

4.6 M<)t so thu thu~t su dvng may tinh (casio) 61

Chu s da 5.1 Khai ni~m ve hlnh da di~n va khoi da di~n 63

5.2 The tich khoi da di~n 64

5.3 M<)t so cong thuc tinh nhanh the' tich khoi chop thuemg gi{tp 68

5.4 M<)t so cong thuc tinh nhanh ve ill di~n 72

5.5 M<)t so cong thuc tinh nhanh ti so the tich 73

chu 6 6.1 Mi{tt 6.2 Mi{tt tr1,1 tron xoay 76

6.3 M.;it cau 77

6.4 M<)t so d~ng to an va cong thuc giai nhanh m.;it non 78

d~ng Chu de 1 Ham s5 8

1.1 Tinh dan di~u cua ham so 8

1.2 Cl;l'c tri ham so 13

1.3 Gia tri I6n nhat, gia tri nho nhat 19

1.4 Duemg ti~m c~n cua do thi ham so 20

1.5 Do thi ham so chua dau gia tri tuy~t doi 21

1.6 Tiep tuyen 22

1.7 Tuang giao do tht 23

Chu de 2 Liiy thira, mii va logarit 28

2.1 Ham so lily thua 28

2.2 Ham so mil 28

2.3, Logarit va ham so logarit 29

2.4 Phuang trlnh mil 30

2.5 Phuang trlnh logarit 32

2.6 Bat phuO'Ilg trlnh mil 33

2.7 Bat phuang trlnh logarit 34

2.8 Bai to an lai suat ngan hang 36

Chu de 3 Nguyen ham- tfch phan- Un.g d\lng tich phan 38

3.1 Bang nguyen ham cua m<)t so ham thuang g~p , 38

3.2 Cac phuO'Ilg phap tinh nguyen ham 38

3.3 Cac phuO'Ilg phap tinh tich phan 41

3.4 Tich phan cac ham so so cap co ban 43

3.5 Ung dvng tich phan 45

3.6 M<)t so phuang phap tinh tich phan ham an 50

3.7 M<)t so thu thu~t su dvng casio 51

Chu de 4 Sd phuc 53

4.1 Kien thuc co ban 53

4.2 Cac phep to an co ban so phuc 53

4 3 T c;tp Q'P 1em 1eu 1en sop uc A h d' A? b' A? d' J:: AI h r 55 4.4 Phuang trlnh b~c hai v6i h~ so thl;l'c 59

4.5 M<)t so bai toan lien quan den min- max so phuc 60

4.6 M<)t so thu thu~t su dvng may tinh (casio) 61

Chu s da 5.1 Khai ni~m ve hlnh da di~n va khoi da di~n 63

5.2 The tich khoi da di~n 64

5.3 M<)t so cong thuc tinh nhanh the' tich khoi chop thuemg gi{tp 68

5.4 M<)t so cong thuc tinh nhanh ve ill di~n 72

5.5 M<)t so cong thuc tinh nhanh ti so the tich 73

chu 6 6.1 Mi{tt 6.2 Mi{tt tr1,1 tron xoay 76

6.3 M.;it cau 77

6.4 M<)t so d~ng to an va cong thuc giai nhanh m.;it non 78 d~ng

6.6 M9t so d~mg toan va cong thuc giaitoan mij.t cau 92

6.7 Tong hqp cac cong thuc dij.c bi~t ve khoi tron xoay 100

Chit de 7 Phuong phap tQa d(} trong khong gian 101

7.1 H~ t9a d9 trong khong gian 101

7.2 Phuotlg trlnh mij.t cau 102

7.3 Mij.t ph~ng 103

' h.:!, 7.4 Duotlg t ang 104

7.5 Cac bai toan tr9ng tam ve t9a d9 106

7.6 Cac bai toan tr9ng tam ve mij.t ph~ng 108

7.7 Cac bai toan tr9ng tam ve duemg th~ng 115

7.8 Cac bai toan tr9ng tam ve mij.t cau 126

Chit de 8 T6 hqp- xac sua't, ca'p so, g6c va khoang each 132

8.1 Hoan vi - chinh hqp - to hqp 132

8.2 Nhi thuc newton · 133

8.3 Xac suat 135

8.4 Cap so c9ng, cap so nhan 137

8.5 Cach xach dinh g6c trong khong gian 138

8.6 Cach xach dinh khoang each trong khong gian 141

A- 300 cau hoi trQng tam 146

1 Ham so va cac bai toan lien quan 146

2 Ham so lily thua, ham so mil va ham so logarit 157

3 Nguyen ham, tich phan va U>ng dvng 161

4 so phuc 166

5 Hlnh h9c khong gian 169

6 Mij.t n6n, mij.t trv, mij.t c'i\l.u 172

7 Phuotlg phap t9a d9 khong gian 174

8 To hqp- xac suat, cap so c9ng- cap so nhan 179

B- Dap an va huong d~n giai chi tie't 182

1 Ham so va cac bai toan lien quan 182

2 Ham so lily thua, ham so mil va ham so logarit 195

3 Nguyen ham, tich phan va U>ng dvng 202

4 so phuc 209

5 Hlnh h9c khong gian 215

6 Mij.t n6n, mij.t trv, mij.t cau 224

7 Phuotlg phap t9a d9 trong khong gian 228

8 To hqp- xac suat, cap so c9ng- cap so nhan 241

A - B(} cau hoi hay -1~ - kh6 hQc tu duy 10 diem 246

B - Dap an va d~n giai chi tie't 262

LAM 1 Chie'n thu~t t6ng on, luy~n de tang 2-3 diem giai do~n nuoc rut 313

2 Bi kip phan b6 thai gian lam de hi~u qua

6.6 M9t so d~mg toan va cong thuc giaitoan mij.t cau 92

6.7 Tong hqp cac cong thuc dij.c bi~t ve khoi tron xoay 100

Chit de 7 Phuong phap tQa d(} trong khong gian 101

7.1 H~ t9a d9 trong khong gian 101

7.2 Phuotlg trlnh mij.t cau 102

7.3 Mij.t ph~ng 103

' h.:!, 7.4 Duotlg t ang 104

7.5 Cac bai toan tr9ng tam ve t9a d9 106

7.6 Cac bai toan tr9ng tam ve mij.t ph~ng 108

7.7 Cac bai toan tr9ng tam ve duemg th~ng 115

7.8 Cac bai toan tr9ng tam ve mij.t cau 126

Chit de 8 T6 hqp- xac sua't, ca'p so, g6c va khoang each 132

8.1 Hoan vi - chinh hqp - to hqp 132

8.2 Nhi thuc newton · 133

8.3 Xac suat 135

8.4 Cap so c9ng, cap so nhan 137

8.5 Cach xach dinh g6c trong khong gian 138

8.6 Cach xach dinh khoang each trong khong gian 141

A-300 cau hoi trQng tam 146

1 Ham so va cac bai toan lien quan 146

2 Ham so lily thua, ham so mil va ham so logarit 157

3 Nguyen ham, tich phan va U>ng dvng 161

4 so phuc 166

5 Hlnh h9c khong gian 169

6 Mij.t n6n, mij.t trv, mij.t c'i\l.u 172

7 Phuotlg phap t9a d9 khong gian 174

8 To hqp- xac suat, cap so c9ng- cap so nhan 179

B- Dap an va huong d~n giai chi tie't 182

1 Ham so va cac bai toan lien quan 182

2 Ham so lily thua, ham so mil va ham so logarit 195

3 Nguyen ham, tich phan va U>ng dvng 202

4 so phuc 209

5 Hlnh h9c khong gian 215

6 Mij.t n6n, mij.t trv, mij.t cau 224

7 Phuotlg phap t9a d9 trong khong gian 228

8 To hqp- xac suat, cap so c9ng- cap so nhan 241

A - B(} cau hoi hay -1~ - kh6 hQc tu duy 10 diem 246

B - Dap an va d~n giai chi tie't 262

LAM 1 Chie'n thu~t t6ng on, luy~n de tang 2-3 diem giai do~n nuoc rut 313 2 Bi kip phan b6 thai gian lam de hi~u

Ky thi THPT Quae gia dang den rat gan, m()t buoc nga~t Ian doi voi mbi si ru, danh dau "nguy~n VQng" th~mh cong dau dffi

L3.m the naa de' but pha rn hQC sinh trung binh tang nhanh len 7, 8 die'm? Lam the naa de d9-t die'm 9, 10 trang ky thi?

Co 5 yeu to quan trQng quyet dinh tai die'm so cua hQC sinh trang giai da9-n nuac rut nay chinh I a:

1 Kien thuc

2 Chien thu~t hQc tang diem 3 Toe d9 lam de, phim x9- de 4 Tam ly phong thi vfmg vang 5 Sv cham chi, kien td

On sai each, Iuoi bieng, chenh mimg se khien lt;I'C hQC cua b9-n mai d~m chan t9-i chb Giai phap tot nhat cha cac b9-n can ngay bay gio chinh la b9 sach I I CAP TOC 789+" tong on thi THPT QG voi 4 mon quan trQng: Taan, Li, Hoa, Anh se giup b9-n:

• T~p trung on dung, du, trung phan kien thuc t~p trung nhat, tranh on Ian man, sa da khong dung mvc tieu diem so

• Cung cap bQ cau hoi ch~t IQc nhat vm nhfrng d9-ng bai muc d@, kho, Ct;I'C kho co kha nang caa xuat hi~n trang de thi

• Tang cu&ng phan x9- de, thu n9-p nhieu phuang phap giai t~t, nham nhanh, daan y giup tang 3 toe d9 lam de

• B6 tUi nhi'eu m~a, tips tranh b§.y hay la kinh nghi~m col-0-2 m anh, chi thu khaa- a khaa di truac

• Va d~c bi~t khong the thieu nhfmg I&i khuyen hfru ich de' vaa phong thi co tam ly thaai mainhat

Day cling Ia b9 sach d~c bi~t co phan chia cac cap de) kien thuc thea rung mvc tieu diem so 7+, 8+, 9+ phu hQp vai mang muon diem thi clia mbi b9-n, giup si ru RUT NGAN thai gian on thi clia mlnh va v§.n hi~u qua caa Ket qua la khi lam tm nhfrng b9 de chuan cau true trang sach, b9-n se hai long vi diem so tang len dang ke

Voi kinh nghi~m nhi'eu nam trang qua tdnh giang d9-y Taan pho thong cling kinh nghi~m giai, phan tich de thi THPT Quae gia, sv nh9-y ben trang phan daan xu huang ra de ket hQ'p n9i dung bam sat rna tr~n de thi chuan se giup cac b9-ll on t~p m9t each hi~u qua nhat

Trang qua trinh bien sa9-TI ch~c ch~n khong tranh het du9c nhfrng sai sot, nham l§.n, mang b9-n dQC thong cam va chia se, dong gop nhfmg th~c m~c de cuon sach haan thi~n han trang nhfrng fan t.ii ban sau

Le Due Thi~u (Chu bien)

Ky thi THPT Quae gia dang den rat gan, m()t buoc nga~t Ian doi voi mbi si ru, danh dau "nguy~n VQng" th~mh cong dau dffi

L3.m the naa de' but pha rn hQC sinh trung binh tang nhanh len 7, 8 die'm? Lam the naa de d9-t die'm 9, 10 trang ky thi?

Co 5 yeu to quan trQng quyet dinh tai die'm so cua hQC sinh trang giai da9-n nuac rut nay chinh I a:

1 Kien thuc

2 Chien thu~t hQc tang diem 3 Toe d9 lam de, phim x9- de 4 Tam ly phong thi vfmg vang 5 Sv cham chi, kien td

On sai each, Iuoi bieng, chenh mimg se khien lt;I'C hQC cua b9-n mai d~m chan t9-i chb Giai phap tot nhat cha cac b9-n can ngay bay gio chinh la b9 sach I I CAP TOC 789+" tong on thi THPT QG voi 4 mon quan trQng: Taan, Li, Hoa, Anh se giup b9-n:

• T~p trung on dung, du, trung phan kien thuc t~p trung nhat, tranh on Ian man, sa da khong dung mvc tieu diem so

• Cung cap bQ cau hoi ch~t IQc nhat vm nhfrng d9-ng bai muc d@, kho, Ct;I'C kho co kha nang caa xuat hi~n trang de thi

• Tang cu&ng phan x9- de, thu n9-p nhieu phuang phap giai t~t, nham nhanh, daan y giup tang 3 toe d9 lam de

• B6 tUi nhi'eu m~a, tips tranh b§.y hay la kinh nghi~m col-0-2 m anh, chi thu khaa- a khaa di truac

• Va d~c bi~t khong the thieu nhfmg I&i khuyen hfru ich de' vaa phong thi co tam ly thaai mainhat

Day cling Ia b9 sach d~c bi~t co phan chia cac cap de) kien thuc thea rung mvc tieu diem so 7+, 8+, 9+ phu hQp vai mang muon diem thi clia mbi b9-n, giup si ru RUT NGAN thai gian on thi clia mlnh va v§.n hi~u qua caa Ket qua la khi lam tm nhfrng b9 de chuan cau true trang sach, b9-n se hai long vi diem so tang len dang ke

Voi kinh nghi~m nhi'eu nam trang qua tdnh giang d9-y Taan pho thong cling kinh nghi~m giai, phan tich de thi THPT Quae gia, sv nh9-y ben trang phan daan xu huang ra de ket hQ'p n9i dung bam sat rna tr~n de thi chuan se giup cac b9-ll on t~p m9t each hi~u qua nhat

Trang qua trinh bien sa9-TI ch~c ch~n khong tranh het du9c nhfrng sai sot, nham l§.n, mang b9-n dQC thong cam va chia se, dong gop nhfmg th~c m~c de cuon sach haan thi~n han trang nhfrng fan t.ii ban sau

Le Due Thi~u (Chu bien)

Cho ham so y = f ( x) co diilo ham tren K

+ Neu f I (X) ;:::: 0 v6i mQi X E K va f I (X) = 0 chi xay ra tiili m()t so hfru hiiln die'm X E K thl ham so y = f ( x) dong bien tren K

+ Neu f I (X) ::; 0 v6i mQi X E K va f I (X) = 0 chi xay ra tiili m()t so hfru hiiln die'm X E K th1 ham so y = f (x) nghich bien tren K

D "'· 01 va1 a m y = - -, h' ax+b ( x :;t: - -dJ kh' ' tinhd 1xet an d'" 1~u t hl f'( ) 0 x = kh" ongxayra en ta ' N."

c>O

Khi g ( x) = 0 co 2 nghi~m phan bi~t x1 ; x2 thl a.g ( x) < 0 Vx E ( x1; X 2 )

{a<O 11::;0

la=O b=O c<O

Tim m de' ham so y = f ( x) dong bien, nghich bien tren khoang (a; b), doiiln [a; b J

Su dvng each xet dau tam thuc b~c 2 (dung cho khi khong co l~p duqc m) Vi~c xet dau f' (X) khi dua duqc ve xet dau cua m()t tam thuc b~c 2 th1 ta dung each xet dau cia m()t tam thuc b~c 2 ben tren de xu ly

Dua ve bai toan min- max (dung cho khi co l~p duqc m)

tht,rc) Co bao nhieu gia tri nguyen cua m de ham soda cho dong bien tren khoang ( 0; +oo)?

Cho ham so y = f ( x) co diilo ham tren K

+ Neu f I (X) ;:::: 0 v6i mQi X E K va f I (X) = 0 chi xay ra tiili m()t so hfru hiiln die'm X E K thl ham so y = f ( x) dong bien tren K

+ Neu f I (X) ::; 0 v6i mQi X E K va f I (X) = 0 chi xay ra tiili m()t so hfru hiiln die'm X E K th1 ham so y = f (x) nghich bien tren K

D "'· 01 va1 a m y = - -, h' ax+b ( x :;t: - -dJ kh' ' tinhd 1xet an d'" 1~u t hl f'( ) 0 x = kh" ongxayra en ta ' N."

c>O

Khi g ( x) = 0 co 2 nghi~m phan bi~t x1 ; x2 thl a.g ( x) < 0 Vx E ( x1; X 2 )

{a<O 11::;0

la=O b=O c<O

Tim m de' ham so y = f ( x) dong bien, nghich bien tren khoang (a; b), doiiln [a; b J

Su dvng each xet dau tam thuc b~c 2 (dung cho khi khong co l~p duqc m) Vi~c xet dau f' (X) khi dua duqc ve xet dau cua m()t tam thuc b~c 2 th1 ta dung each xet dau cia m()t tam thuc b~c 2 ben tren de xu ly

Dua ve bai toan min- max (dung cho khi co l~p duqc m)

Di'eu ki~n xac dinh: X * m 4-m2 Ta c6 f' ( x) = 2

(x-m)

Ham soda cho dong bien tren khoang ( o; +oo)

<=> 4 m2 lm ~ (O;+oo)

f'(x)= - 2 >0 VxE(O;+oo) <=> { <=> <=> -2 < m s 0

4-m2

>0 -2<m<2 (x-m)

cua m th6a man cau bai toan Biet t?p hqp tat ca cac gia tri tht,l'C cua tham so m de? ham so

y = 2x3 -3(2m+l)x' +6m(m + l)x+ 1 dOng bie'n tri'n khoang (2;+oo) lit ( -oo;~ J~ lit phan sO toi gian Khi d6 2a + b c6 gia tri b&ng

T?p xac dinh D = 1R Ta c6 y' = 6x2

Ham so luon dong bien tren 1R => y' ~ o, v x E 1R

<=>~so

<=> (2m+1r -4m(m+1) so <=> 1 s 0 (L)

Phuong tdnh y' = 0 c6 hai nghi~m phan bi~t th6a man x1 < x2 s 2 <=> x1 - 2 < x2 - 2 s 0

+ 14mx+5 s 0, Vx ~ 1 <=> g(x) = -~ m (1) +14x

C6 g'(x) = 2 > 0, Vx E [ 1; +oo), suy ra ming(x) = g(1) =

Di'eu ki~n xac dinh: X * m 4-m2 Ta c6 f' ( x) = 2

(x-m)

Ham soda cho dong bien tren khoang ( o; +oo)

<=> 4 m2 lm ~ (O;+oo)

f'(x)= - 2 >0 VxE(O;+oo) <=> { <=> <=> -2 < m s 0

4-m2

>0 -2<m<2 (x-m)

cua m th6a man cau bai toan Biet t?p hqp tat ca cac gia tri tht,l'C cua tham so m de? ham so

y = 2x3 -3(2m+l)x' +6m(m + l)x+ 1 dOng bie'n tri'n khoang (2;+oo) lit ( -oo;~ J~ lit phan sO toi gian Khi d6 2a + b c6 gia tri b&ng

T?p xac dinh D = 1R Ta c6 y' = 6x2

Ham so luon dong bien tren 1R => y' ~ o, v x E 1R

<=>~so

<=> (2m+1r -4m(m+1) so <=> 1 s 0 (L)

Phuong tdnh y' = 0 c6 hai nghi~m phan bi~t th6a man x1 < x2 s 2 <=> x1 - 2 < x2 - 2 s 0

+ 14mx+5 s 0, Vx ~ 1 <=> g(x) = -~ m (1) +14x

C6 g'(x) = 2 > 0, Vx E [ 1; +oo), suy ra ming(x) = g(1) =

Xet Sl,f tuang giao cua do thi ham so so y = £' (X) va duemg th~ng d : y = 1 (nhu hlnh ve ben duoi)

Xet Sl,f tuang giao cua do thi ham so so y = £' (X) va duemg th~ng d : y = 1 (nhu hlnh ve ben duoi)

Dl,fa vao do thj ta co g' (X)< 0 -1 <X< 2 => g ( -1) > g ( 1) > g ( 2)

Cho ham so y = £ (X) co dl;lO ham tren ~ thoa man f ( 2) = f ( -2) = 0 Va dO thi CUa ham SO Y = f 1

(X) CO d1;1ng nhU hlnh ben Ham so y = ( f (X) r nghich bien tren khoang nao trong d.c khoang sau ?

( -1; n ( -1; 1) (1;2 )

Ta co £I (x) = 0 <=> x = 1; x = ±2; £ ( 2) = £ ( -2) = 0 Ta co bang bien thien:

=> f(x) < O;'v'x * ±2

Taco g'(x) = 2f'(x).f(x) Xet g'(x) < 0 <=> f'(x).f(x) < 0 <=> {f'((x)) > 0 <=> [x < -2 £ x <0 1<x<2 Suy ra ham so g (x) nghich bien tren cac khoang ( -oo; -2) I ( 1; 2)

Cho ham so y = £ ( x) Do thi ham so y = f' ( x) nhu hinh ben Ham so g (X) = £ (13- xl) dong bien tren khoang nao du6i day?

( -oo;1) ( 2;4 )

Dung MODE 7 nh~p ham; quan sat bang gia tri ham so; khoang nElO bang gia tri luon

tang thl ham so dong bien tren khoang do; khoang nao bang gia tri luon giam thl ham so nghich

Cho ham so y = £ (X) co dl;lO ham tren ~ thoa man f ( 2) = f ( -2) = 0 Va dO thi CUa ham SO Y = f 1

(X) CO d1;1ng nhU hlnh ben Ham so y = ( f (X) r nghich bien tren khoang nao trong d.c khoang sau ?

( -1; n ( -1; 1) (1;2 )

Ta co £I (x) = 0 <=> x = 1; x = ±2; £ ( 2) = £ ( -2) = 0 Ta co bang bien thien:

=> f(x) < O;'v'x * ±2

Taco g'(x) = 2f'(x).f(x) Xet g'(x) < 0 <=> f'(x).f(x) < 0 <=> {f'((x)) > 0 <=> [x < -2 £ x <0 1<x<2 Suy ra ham so g (x) nghich bien tren cac khoang ( -oo; -2) I ( 1; 2)

Cho ham so y = £ ( x) Do thi ham so y = f' ( x) nhu hinh ben Ham so g (X) = £ (13- xl) dong bien tren khoang nao du6i day?

( -oo;1) ( 2;4 )

Dung MODE 7 nh~p ham; quan sat bang gia tri ham so; khoang nElO bang gia tri luon

tang thl ham so dong bien tren khoang do; khoang nao bang gia tri luon giam thl ham so nghich

Su dl,lng d9-0 ham t9-i 1 die'm neu Xo) 0; Xo E (a; b) thl ham so khong dong bien tren khming (a; b); neu f' ( X0 ) > 0; X0 E (a; b) thl ham SO khong nghich bien tren khm\ng (a; b)

Ham so y = ~ 8 + 2x- x2

dong bien tren khoimg nao sau day? ( 1; + 00) ( 1; 4)

chQn START: -5; END: 4: STEP: 0.5

thi

-0 5

an ghi tri f (X) khong xac dinh ho?c giam V~y chQn ( -2; 1)

Dung CALC de thu cac gia tri trong khoang can xet

Bam CALC gia tri X ta b6 qua; bam dau "=" va gan A gia tri 1,1 ta thay ket qua am tuc mQi khoang chua X = 1,1 thl khong dong bien V~y lo9-i dap an A va dap an B

Con dap an C va dap an D thl tiep tl,lc tinh d9-o ham t9-i 1 diem thu()c ( -oo; 1) rna khong thu()c ( -2;1) vi d1,1 nhu x = -5 ta duqc ket qua la Math ERRO V~y

uoc tinh Tinh d9-o ham t9-i die'm x=1,1

Tinh d9-o ham t9-i die'm x=-5

c.:snce J: Giot.o

T, 1111 tat ca cac gm " ' ' , , ., tq h t vc cua , t h am so , , m dA'h' e am so y , , = cosx-2 ng c hihb'"' 1en tren A cosx-m

D :;;: h"" e t ay va1 m = t , 2 hl y = cosx- 2 = 1 A h' nen am so ""khA ong d an d'A 1~u tren A kh ' oang (o ; -nJ 09-1 L · C

chQn START: -5; END: 4: STEP: 0.5

thi

-0 5

an ghi tri f (X) khong xac dinh ho?c giam V~y chQn ( -2; 1)

Dung CALC de thu cac gia tri trong khoang can xet

Bam CALC gia tri X ta b6 qua; bam dau "=" va gan A gia tri 1,1 ta thay ket qua am tuc mQi khoang chua X = 1,1 thl khong dong bien V~y lo9-i dap an A va dap an B

Con dap an C va dap an D thl tiep tl,lc tinh d9-o ham t9-i 1 diem thu()c ( -oo; 1) rna khong thu()c ( -2;1) vi d1,1 nhu x = -5 ta duqc ket qua la Math ERRO V~y

uoc tinh Tinh d9-o ham t9-i die'm x=1,1

Tinh d9-o ham t9-i die'm x=-5

c.:snce J: Giot.o

T, 1111 tat ca cac gm " ' ' , , ., tq h t vc cua , t h am so , , m dA'h' e am so y , , = cosx-2 ng c hihb'"' 1en tren A cosx-m

D :;;: h"" e t ay va1 m = t , 2 hl y = cosx- 2 = 1 A h' nen am so ""khA ong d an d'A 1~u tren A kh ' oang (o ; -nJ 09-1 L · C

Thay cac gia tri cua F (X) giam dan vay chQn dap an chua m = 1 =>

+ Buac 1: Tim tap xac dinh Tfnh f'(x)

+ Buac 2: Tim cac diem xi rna tc;ti d6 f' (X)= 0 ho~c f' (X) khong xac dinh

+ Buac 3: Lap bang bien thien

+ Buac 4: Neu f' ( x) doi dau khi di qua xi thi ham so dc;tt cvc tri tc;ti xi

+ Buac 1: Tim tap xac dinh Tim f'(x)

+ Buac 2: Tim cac nghi~m xi (i = li2i ) cua phuong trinh f'(x) = 0

x1 < < 0 P > 0

£) x1 < a < x2 <=> ( x1 - a) ( x2 - a) < 0 <=> x1 .x2 - a ( x1 + x2 ) + a 2 < 0

( x1 -a) ( x2 a) > 0 {x1 .x2 - a ( x1 + x2 ) + a 2 > 0 g) < < a <=> <=>

x1 + x2 < 2a x1 + x2 < 2a , , {( X 1 - a) ( x2 - a) > 0 , , {x1 .x2 - a ( x1 + x2 ) + a 2

> 0 h) a < x1 < 'r-7 'r-7

x + x > 2a x + x > 2a

Thay cac gia tri cua F (X) giam dan vay chQn dap an chua m = 1 =>

+ Buac 1: Tim tap xac dinh Tfnh f'(x)

+ Buac 2: Tim cac diem xi rna tc;ti d6 f' (X)= 0 ho~c f' (X) khong xac dinh

+ Buac 3: Lap bang bien thien

+ Buac 4: Neu f' ( x) doi dau khi di qua xi thi ham so dc;tt cvc tri tc;ti xi

+ Buac 1: Tim tap xac dinh Tim f'(x)

+ Buac 2: Tim cac nghi~m xi (i = li2i ) cua phuong trinh f'(x) = 0

x1 < < 0 P > 0

£) x1 < a < x2 <=> ( x1 - a) ( x2 - a) < 0 <=> x1 .x2 - a ( x1 + x2 ) + a 2 < 0

( x1 -a) ( x2 a) > 0 {x1 .x2 - a ( x1 + x2 ) + a 2 > 0 g) < < a <=> <=>

x1 + x2 < 2a x1 + x2 < 2a , , {( X 1 - a) ( x2 - a) > 0 , , {x1 .x2 - a ( x1 + x2 ) + a 2

> 0 h) a < x1 < 'r-7 'r-7

x + x > 2a x + x > 2a

+ A, B n~m cimg ve 1 phia doi v6i trt.tc Oy <=> y' = 0 co hai nghi~m phan bi~t cling dau + A, B n~m cling ve 2 phia doi v6i tn,1.c Oy <=> y' = co hai nghi~m trcli dau

+A, B n~m cling ve 1 phia doi v6i trt,lC Ox ¢:> y' = 0 co hai nghi~m phan bi~t va y CD .y cr > 0

[1\ '> 0 + A, B n~m cling ve phia tren doi v6i tr1,1.c Ox <=> y ~ .y cr > 0

Yco+Ycr>O

+A, B n~m cimg ve phia du6i doi v6i trt.tc Ox <=> y~.y cr > 0

[1\ '> 0 Yco+Ycr<O

A B ~ "' 2 hi d"'' ' 0 {1\y, > 0 + , nam ve p a 01 vm tn,1.c x <=>

Yco·Ycr <0

(ap dvng khi khong nham duqc nghi~m va viet duqc phuang trlnh duemg th~ng di qua hai diem Cl,fC tri cua do thi ham so)

+ A, B n~m ve 2 phia doi v6i tn.tc Ox

<=> do thi c~t trvc Ox t;;1.i 3 diem phan bi~t

<=> phuang trlnh hoanh d9 giao ~iem f(x) = 0 co 3 nghi~m phan bi~t (ap dl,IDg khi nham duqc nghi~m)

Gici su ham sob~c ba y =£(X)= ax3 + bx2 + cx+d( a* 0) co hai die'm Cl,fC tri la xl;x2 Khi do, tht,rc hi~n phep chia f ( x) cho £' ( x) ta duQ'c : f ( x) = Q ( x) £' ( x) +Ax+ B

, , {y 1 = f ( x1 ) = Ax1 + B Do do, ta co: _ ( ) _

y 2 - f x2 - Ax2 + B

Suy ra, cac diem (x1;y1),(x2;y2) n~m tren duemg th~ng y =Ax+ B

+ Ham so co m9t ct,rc tri <=> ab ;;::: 0 + Ham so co ba ct,rc tri <=> ab < 0

{a>O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC tieu ¢:> b ;;::: 0

{a<O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC dgi ¢:> b s 0

{a>O + Ham so co hai Cl,fC tieu va m9t Cl,l'C d;;li <=> b < 0

{a<O + Ham so co m9t Cl,fC tieu va hai Cl,l'C d;;li <=> b > 0

GUtsuhamso y=ax4+bx2+c co 3 cvctri: A(O;c),B(-~-b ;-~J,c(~-b ;-~J

+ A, B n~m cimg ve 1 phia doi v6i trt.tc Oy <=> y' = 0 co hai nghi~m phan bi~t cling dau + A, B n~m cling ve 2 phia doi v6i tn,1.c Oy <=> y' = co hai nghi~m trcli dau

+A, B n~m cling ve 1 phia doi v6i trt,lC Ox ¢:> y' = 0 co hai nghi~m phan bi~t va y CD .y cr > 0

[1\ '> 0 + A, B n~m cling ve phia tren doi v6i tr1,1.c Ox <=> y ~ .y cr > 0

Yco+Ycr>O

+A, B n~m cimg ve phia du6i doi v6i trt.tc Ox <=> y~.y cr > 0

[1\ '> 0 Yco+Ycr<O

A B ~ "' 2 hi d"'' ' 0 {1\y, > 0 + , nam ve p a 01 vm tn,1.c x <=>

Yco·Ycr <0

(ap dvng khi khong nham duqc nghi~m va viet duqc phuang trlnh duemg th~ng di qua hai diem Cl,fC tri cua do thi ham so)

+ A, B n~m ve 2 phia doi v6i tn.tc Ox

<=> do thi c~t trvc Ox t;;1.i 3 diem phan bi~t

<=> phuang trlnh hoanh d9 giao ~iem f(x) = 0 co 3 nghi~m phan bi~t (ap dl,IDg khi nham duqc nghi~m)

Gici su ham sob~c ba y =£(X)= ax3 + bx2 + cx+d( a* 0) co hai die'm Cl,fC tri la xl;x2 Khi do, tht,rc hi~n phep chia f ( x) cho £' ( x) ta duQ'c : f ( x) = Q ( x) £' ( x) +Ax+ B

, , {y 1 = f ( x1 ) = Ax1 + B Do do, ta co: _ ( ) _

y 2 - f x2 - Ax2 + B

Suy ra, cac diem (x1;y1),(x2;y2) n~m tren duemg th~ng y =Ax+ B

+ Ham so co m9t ct,rc tri <=> ab ;;::: 0 + Ham so co ba ct,rc tri <=> ab < 0

{a>O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC tieu ¢:> b ;;::: 0

{a<O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC dgi ¢:> b s 0

{a>O + Ham so co hai Cl,fC tieu va m9t Cl,l'C d;;li <=> b < 0

{a<O + Ham so co m9t Cl,fC tieu va hai Cl,l'C d;;li <=> b > 0

GUtsuhamso y=ax4+bx2+c co 3 cvctri: A(O;c),B(-~-b ;-~J,c(~-b ;-~J

- - a -b3

DiH a=BAC :::>coe-=-

Tam giac ABC vuong can t~i A

~~·~~~~···~···"~···~~~·~~····~·~

Tam giac ABC c6 3 g6c nhQn

Tr1;1c hoanh chia tam giac ABC thanh hai haiphan c6 di~n tich nhau

Tam giac ABC c6 diem c\].'c tri each deu Ox Do thi ham so (C): y = ax4

+ bx2

+ c c~t tr1;1c Ox tCJ.i 4 diem phan th~mh

Hlnh ph~ng gioi h9.n boi do thi

phan tren va phan duoi ox b~ng nhau

b2 =8ac b2 = 100 ac

Tam giac ABC vuong can t~i A

~~·~~~~···~···"~···~~~·~~····~·~

Tam giac ABC c6 3 g6c nhQn

Tr1;1c hoanh chia tam giac ABC thanh hai haiphan c6 di~n tich nhau

Tam giac ABC c6 diem c\].'c tri each deu Ox Do thi ham so (C): y = ax4

+ bx2

+ c c~t tr1;1c Ox tCJ.i 4 diem phan th~mh

Hlnh ph~ng gioi h9.n boi do thi

phan tren va phan duoi ox b~ng nhau

b2 =8ac b2 = 100 ac

9

X

Tlm m de do thi ham so y = x - mx + co ba diem cvc tri Iap thanh m()t tam ghic

Ba diem Cl,l'C tri Iap thanh m()t tam giac vuong thl b3

= -8a <=> -m3 = -8 <=> m = 2 So gia tri nguyen cua tham so mde do thi cua ham so y = x4

- 2 ( m + 1) x2 + m 2 co ba

Di'eu ki~n de' do thi ham trung phuong y = ax4

+ bx2 + c co ba diem cvc tri la ab < 0 <=> m > -1 Khi do ba diem Cl,l'C tri Iap thanh tam giac vuong can khi b3

+ 8a = 0 <=> -8( m + 1 t + 8 = 0 <=> m = 0 Cho ham so y = 3x4 - 2mx2 +2m+ m 4 • Tlm tat dt cac gia tri cua m de' do thi ham soda cho co ba diem cvc tri t9-o thanh tam giac co di~n tich b~ng 3

Di'eu ki~n de' do thi ham trung phuong y = ax4 + bx2

+ c co ba die'm cvc tri la ab < 0 <=> m > 0 Khi do ba die'm cvc tri t9-o thanh tam giac co di~n tich b~ng 3 khi

32.33.32 +(-2m r = 0 <=> m = 3

So gia tri thvc cua tham so m de do thi ham so y = x4 - 2mx2

+ m -1 co ba diem Cl,l'C tri t9-o thanh m()t tam giac co ban kfnh du&ng tron ngo9-i tiep b~ng 1 la ?

Di'eu ki~n de ham so co 3 diem cvc tri la: 1( -2m)< o <=> m > o

Ap dvng cong thuc ba diem Cl,l'C tri ham hung phuong tgo thanh m()t tam giac co ban kinh du&ng , '"" , b' k'nh R b3 -8a 1 (-2mf -8

tron ngo9.1 hep co an 1 = -~-~ - <::> = ( )

- 2 ( m + 1) x2 + m 2 co ba

Di'eu ki~n de' do thi ham trung phuong y = ax4

+ bx2 + c co ba diem cvc tri la ab < 0 <=> m > -1 Khi do ba diem Cl,l'C tri Iap thanh tam giac vuong can khi b3

+ 8a = 0 <=> -8( m + 1 t + 8 = 0 <=> m = 0 Cho ham so y = 3x4 - 2mx2 +2m+ m 4 • Tlm tat dt cac gia tri cua m de' do thi ham soda cho co ba diem cvc tri t9-o thanh tam giac co di~n tich b~ng 3

Di'eu ki~n de' do thi ham trung phuong y = ax4 + bx2

+ c co ba die'm cvc tri la ab < 0 <=> m > 0 Khi do ba die'm cvc tri t9-o thanh tam giac co di~n tich b~ng 3 khi

32.33.32 +(-2m r = 0 <=> m = 3

So gia tri thvc cua tham so m de do thi ham so y = x4 - 2mx2

+ m -1 co ba diem Cl,l'C tri t9-o thanh m()t tam giac co ban kfnh du&ng tron ngo9-i tiep b~ng 1 la ?

Di'eu ki~n de ham so co 3 diem cvc tri la: 1( -2m)< o <=> m > o

Ap dvng cong thuc ba diem Cl,l'C tri ham hung phuong tgo thanh m()t tam giac co ban kinh du&ng , '"" , b' k'nh R b3 -8a 1 (-2mf -8

tron ngo9.1 hep co an 1 = -~-~ - <::> = ( )

Co bao nhieu gia tri nguyen cua tham so m E [ -10; 10 J de ham so

y = lmx3 - 3mx2

+(3m- 2)x + 2- ml co 5 diem cvc tri?

Xet ham so f ( x) = mx3 - 3mx2

+ (3m-2) x +2-m

Taco: mx3-3mx2

+(3m-2)x+2-m=O <=>[x=2l ( )' mx -2mx+m-2 =0 1

Yeu cau bai toan ¢::> phuang trlnh f (X) = 0 co ba nghi~m phan bi~t ¢::> phuang trlnh ( 1) co hai

{m2

-m(m-2) > 0 nghi~m phan bi~t khac 1 <=>

m-2m+m-2 =F 0 Vl m nguyen va mE [ -10;10] nen mE {1;2; ;10}

So gia tri nguyen cua tham so m de' ham so y = lx3 - 3x2

- 9x- 5 + m 2

1 co 5 diem

Ve do thi ham so y = x3 - 3x2

- 9x-5 nhu hlnh ben du6i

Ham so y = f ( x) co 2 diem cvc tri => y = f ( x) + m 2

cling luon co 2 diem cvc tri Do do yeu cau bai to an ¢::> so giao die'm cua do thi y = f (X) + m 2 v6i tn;tc hoanh la 3

32 2 0 {m2 < 32 {-m < m < .J32

<=> - < -m < <=> <=> m:t=O m=F-0

V6i mE z =>mE {-5;-4;-3;-2;-1;1;2;3;4;5} V~y colO gia tri nguyen m thoa man Cho ham so f (X) = x3

+(3m- 2)x + 2- ml co 5 diem cvc tri?

Xet ham so f ( x) = mx3 -3mx2

+ (3m-2) x +2-m

Taco: mx3-3mx2

+(3m-2)x+2-m=O <=>[x=2l ( )' mx -2mx+m-2 =0 1

Yeu cau bai toan ¢::> phuang trlnh f (X) = 0 co ba nghi~m phan bi~t ¢::> phuang trlnh ( 1) co hai

{m2

-m(m-2) > 0 nghi~m phan bi~t khac 1 <=>

m-2m+m-2 =F 0 Vl m nguyen va mE [ -10;10] nen mE {1;2; ;10}

So gia tri nguyen cua tham so m de' ham so y = lx3 - 3x2

- 9x- 5 + m 2

1 co 5 diem

Ve do thi ham so y = x3 - 3x2

- 9x-5 nhu hlnh ben du6i

Ham so y = f ( x) co 2 diem cvc tri => y = f ( x) + m 2

cling luon co 2 diem cvc tri Do do yeu cau bai to an ¢::> so giao die'm cua do thi y = f (X) + m 2 v6i tn;tc hoanh la 3

32 2 0 {m2 < 32 {-m < m < .J32

<=> - < -m < <=> <=> m:t=O m=F-0

V6i mE z =>mE {-5;-4;-3;-2;-1;1;2;3;4;5} V~y colO gia tri nguyen m thoa man Cho ham so f (X) = x3

(2m-1) -3(2-m)>O 1.>0

~ S > 0 ~ 2 (2m -1) > 0

P>O 2-m 3 - - > 0

g' ( x) = 0 ~ , ( 3 2) _ <:::> x + 3x = a, (a < 0) £ x + 3x - o x + x = , < < 3 3 2 b (o b 4)

x3 + 3x2

Dl,l'a vao bang bien thlen tren ta thay: Phuong trlnh x3

+ 3x2

= a, (a > 0) c6 m{)t nghi~m don x = x1 < -2 Phuong trlnh x3

(2m-1) -3(2-m)>O 1.>0

~ S > 0 ~ 2 (2m -1) > 0

P>O 2-m 3 - - > 0

g' ( x) = 0 ~ , ( 3 2) _ <:::> x + 3x = a, (a < 0) £ x + 3x - o x + x = , < < 3 3 2 b (o b 4)

x3 + 3x2

Dl,l'a vao bang bien thlen tren ta thay: Phuong trlnh x3

+ 3x2

= a, (a > 0) c6 m{)t nghi~m don x = x1 < -2 Phuong trlnh x3

0

+ Cach 1: Dung dgo ham tgi die'm Ian c~n x0

- N eu f I ( Xo + 011) .£' ( xo - 0 I 1) < 0 thl ham so dgt eve tri tgi X = Xo

N {f1

(x0+011)>0 , - Neu 1 ) => x0 Ia eve dgi

f xo -011 < 0

N {f'(x0+011)<0 , '"' - N eu 1 ( ) ::::> x0 Ia eve tieu

+ Buac 2: L~p bang bien thien va ru do suy ra gia tri Ion nhatl gia tri nh6 nhat cua ham so

+ Buac 1:

* Ham so da cho y = f (X) xac dinh va lien h,lc tren dogn [a; b J

* T1m cite die'm XII x21 '"I xn tren khoang (a; b) I tgi do f1

+ Cach 1: Dung dgo ham tgi die'm Ian c~n x0

- N eu f I ( Xo + 011) .£' ( xo - 0 I 1) < 0 thl ham so dgt eve tri tgi X = Xo

N {f1

(x0+011)>0 , - Neu 1 ) => x0 Ia eve dgi

f xo -011 < 0

N {f'(x0+011)<0 , '"' - N eu 1 ( ) ::::> x0 Ia eve tieu

+ Buac 2: L~p bang bien thien va ru do suy ra gia tri Ion nhatl gia tri nh6 nhat cua ham so

+ Buac 1:

* Ham so da cho y = f (X) xac dinh va lien h,lc tren dogn [a; b J

* T1m cite die'm XII x21 '"I xn tren khoang (a; b) I tgi do f1

Bcli toan : Tim min, max cua ham so y = f ( x) tren khoang (a; b), [a; b J

Phuong phap: Su dvng nh?p ham so f ( x); ch<;>n Start: a; End: b; Step: b 1~ a ho~c b 2~ a Quan sat bang gia tri f (X) I gia tri 16n nhat 1a max (xap xi max) I gia tri nh6 nhat 1a min (xap xi min)

Chu y: Neu de bai cho cac khoang (a; +oo) I ( -00; b) I ( -00; +oo) th1 ta thay 00 la cac so C\1 the nhu 10,20 de fim min, max tren rung khoang nh6 va thao tac nhu thong thuang

Gia tri nh6 nhat cua ham so f (X) = x3 -3x + 1 tren do!itn [ -3; 3 J b~ng

Cho ham so y = f(x) xac dinh tren m<)t khoang vo hq.n (la khoang dq.ng (a; +oo) I ( -oo; b) ho~c ( -oo; +oo) ) Duang th~ng y =Yo 1a duang ti~m c~n ngang (hay ti~m c?n ngang) cua do thi ham so y = f( neu it nhat m<)t trong cac di'eu ki~n sau duqc tho a man:

lim f ( x) =Yo, lim f ( x) =Yo

TA? ong so uong Nd ' ti~m •A cq.n ung va A d' ' oA h~m cq.n ngang cua o t am so A ? d"' hih' N y = 2x+3 1' r: -? a -v4-x2

Bcli toan : Tim min, max cua ham so y = f ( x) tren khoang (a; b), [a; b J

Phuong phap: Su dvng nh?p ham so f ( x); ch<;>n Start: a; End: b; Step: b 1~ a ho~c b 2~ a Quan sat bang gia tri f (X) I gia tri 16n nhat 1a max (xap xi max) I gia tri nh6 nhat 1a min (xap xi min)

Chu y: Neu de bai cho cac khoang (a; +oo) I ( -00; b) I ( -00; +oo) th1 ta thay 00 la cac so C\1 the nhu 10,20 de fim min, max tren rung khoang nh6 va thao tac nhu thong thuang

Gia tri nh6 nhat cua ham so f (X) = x3 -3x + 1 tren do!itn [ -3; 3 J b~ng

Cho ham so y = f(x) xac dinh tren m<)t khoang vo hq.n (la khoang dq.ng (a; +oo) I ( -oo; b) ho~c ( -oo; +oo) ) Duang th~ng y =Yo 1a duang ti~m c~n ngang (hay ti~m c?n ngang) cua do thi ham so y = f( neu it nhat m<)t trong cac di'eu ki~n sau duqc tho a man:

lim f ( x) =Yo, lim f ( x) =Yo

TA? ong so uong Nd ' ti~m •A cq.n ung va A d' ' oA h~m cq.n ngang cua o t am so A ? d"' hih' N y = 2x+3 1' r: -? a -v4-x2

de tim ti~m c~n ngang

CALC v6i x=-106 de tim ti~m c~n ngang

V~y do thi ham soda cho chi c6 2 ti~m c~n ngang, khong c6 ti~m c~n dtffig

CALC v6i x = 106 ti~m c~n ngang

CALC v6i 1

de tim ti~m c~n ngang

CALC v6i x=-106 de tim ti~m c~n ngang

V~y do thi ham soda cho chi c6 2 ti~m c~n ngang, khong c6 ti~m c~n dtffig

CALC v6i x = 106 ti~m c~n ngang

CALC v6i 1

* Cach ve:

+ Gift nguyen ph'an do thi ben phai Oy cua do thi (c) : y = f (X)

+ B6 phan do thi ben trcii Oy cua (c) I lay doi x{mg phan do thi dUQ'C gift qua Oy Vedothihamso (c'):y=lxl3 -31xl+l.Tac6dothiham y=f(x)=x3 -3x+1 (c)

+ B6 phan do thi cua (c) ben trai Oy I gift nguyen (c) ben phai Oy + Lay doi x{mg phan do thi dUQ'C gift qua Oy

(c'):y=lxl3 -31xl+1

Ve do thi (c'): y = l£(x)l Ta c6: y = 1£ ( x )I = {£ ( x) khi f ( x) 2 0 -f (X) khi f (X) < 0

* Cach ve ( c') ru (c) :

+ Gift nguyen phan do thi phia tren Ox ~ua do thi (C): y = f (X)

+ B6 phan do thi phia du6i Ox cua (C), lay doi x-ung phan do thi bi b6 qua Ox

Ve do thi y = lx3 - 3x + 11 y (c) : Y = x3 _ 3x + 1 + Ve do thi ham soy= x3

-3x+1( C)

+ B6 phan do thi cua (C) du6i Ox, gift nguyen (C) phia tren Ox

+ Lay doi x{mg phan do thi bi b6 qua Ox

v6i dfilng: y = lf(lxl)i ta Ian luQ't bien doi 2 do thi y = f(lxl) va y = 1£(x )I

Cho ham so y = f ( x) , c6 do thi (C)

- 3x + 11

Tiep tuyen cia do thi (C) tiili die'm M0 ( x0 y 0 e (C) c6 diiing: y = y' ( x0 ) ( x-x0 ) + y 0 •

Mo ( Xo; y 0) la tiep diem; k = f I ( Xo ) la h~ so g6c cua tiep tuyen Cho hai ham so (c) : y = f (X) va ( c I) : y = g (X)

DO !hi (C) vii ( C') tie'p xUc nhau khi vi> chi khi h~ phuang trinh: { : \ : \ : ~~( ~) c6 nghi~m

* Cach ve:

+ Gift nguyen ph'an do thi ben phai Oy cua do thi (c) : y = f (X)

+ B6 phan do thi ben trcii Oy cua (c) I lay doi x{mg phan do thi dUQ'C gift qua Oy Vedothihamso (c'):y=lxl3 -31xl+l.Tac6dothiham y=f(x)=x3 -3x+1 (c)

+ B6 phan do thi cua (c) ben trai Oy I gift nguyen (c) ben phai Oy + Lay doi x{mg phan do thi dUQ'C gift qua Oy

(c'):y=lxl3 -31xl+1

Ve do thi (c'): y = l£(x)l Ta c6: y = 1£ ( x )I = {£ ( x) khi f ( x) 2 0 -f (X) khi f (X) < 0

* Cach ve ( c') ru (c) :

+ Gift nguyen phan do thi phia tren Ox ~ua do thi (C): y = f (X)

+ B6 phan do thi phia du6i Ox cua (C), lay doi x-ung phan do thi bi b6 qua Ox

Ve do thi y = lx3 - 3x + 11 y (c) : Y = x3 _ 3x + 1 + Ve do thi ham soy= x3

-3x+1( C)

+ B6 phan do thi cua (C) du6i Ox, gift nguyen (C) phia tren Ox

+ Lay doi x{mg phan do thi bi b6 qua Ox

v6i dfilng: y = lf(lxl)i ta Ian luQ't bien doi 2 do thi y = f(lxl) va y = 1£(x )I

Cho ham so y = f ( x) , c6 do thi (C)

- 3x + 11

Tiep tuyen cia do thi (C) tiili die'm M0 ( x0y 0 e (C) c6 diiing: y = y' ( x0 ) ( x-x0 ) + y 0 •

Mo ( Xo; y 0) la tiep diem; k = f I ( Xo ) la h~ so g6c cua tiep tuyen Cho hai ham so (c) : y = f (X) va ( c I) : y = g (X)

DO !hi (C) vii ( C') tie'p xUc nhau khi vi> chi khi h~ phuang trinh: { : \ : \ : ~~( ~) c6 nghi~m

Bai to an: Viet phuang trlnh tiep tuyen t~i diem M ( x0 y 0 )

Phuong trlnh tiep tuyen co d~ng d: y = kx + m; ta se tlm k, m nhu sau: + B1: Tlm h~ so goc tiep tuyen k = y' ( x0) = ~(£ ( x ))I _

vay phuang trinh tiep tuyen la: y = 3x- 4

3

Ansx(-1)

Tiep tuyen cua do thi (c) : y = -4x3 + 3x + 1 di qua diem A ( -1; 2) co pmxanLg

Ta co phuang trlnh: -4x3 + 3x + 1 = -9x + 7 co 3 nghi~m phan bi~t ,lo~i A Ta co phuang trlnh: -4x3 + 3x + 1 = -9x ~ 11 co 1 nghi~m dan, lo~i B Ta co phuang trlnh: -4x3 + 3x + 1 = -9x + 11 co 1 nghi~m dan, lo~i C Ta co phuang trlnh: -4x3

+ 3x + 1 = -9x -7 co 2 nghi~m phan bi~t; me la trong 2 nghi~m do se co 1 nghi~m kep nen y = -9x -7la tiep tuyen

Cho ham so y = f(x) co do thi (C1 va y = g(x) co do thi (C2)

Phuong trlnh hoanh d9 giao diem cia (C1 va (C2 ) la f(x) = g(x) (1) Khi d6: + So giao die'm cua ( C1) va ( C2) bilng voi so nghi~m cua phuang trlnh ( 1) + N ghi~m cua phuang trlnh ( 1) chinh la hoanh d9 Xo cua giao diem +Diem M(x;f(x lagiaodiemcua (C va (C

Bai to an: Viet phuang trlnh tiep tuyen t~i diem M ( x0 y 0 )

Phuong trlnh tiep tuyen co d~ng d: y = kx + m; ta se tlm k, m nhu sau: + B1: Tlm h~ so goc tiep tuyen k = y' ( x0) = ~(£ ( x ))I _

vay phuang trinh tiep tuyen la: y = 3x- 4

3

Ansx(-1)

Tiep tuyen cua do thi (c) : y = -4x3 + 3x + 1 di qua diem A ( -1; 2) co pmxanLg

Ta co phuang trlnh: -4x3 + 3x + 1 = -9x + 7 co 3 nghi~m phan bi~t ,lo~i A Ta co phuang trlnh: -4x3 + 3x + 1 = -9x ~ 11 co 1 nghi~m dan, lo~i B Ta co phuang trlnh: -4x3 + 3x + 1 = -9x + 11 co 1 nghi~m dan, lo~i C Ta co phuang trlnh: -4x3

+ 3x + 1 = -9x -7 co 2 nghi~m phan bi~t; me la trong 2 nghi~m do se co 1 nghi~m kep nen y = -9x -7la tiep tuyen

Cho ham so y = f(x) co do thi (C1 va y = g(x) co do thi (C2)

Phuong trlnh hoanh d9 giao diem cia (C1 va (C2 ) la f(x) = g(x) (1) Khi d6: + So giao die'm cua ( C1) va ( C2) bilng voi so nghi~m cua phuang trlnh ( 1) + N ghi~m cua phuang trlnh ( 1) chinh la hoanh d9 Xo cua giao diem +Diem M(x;f(x lagiaodiemcua (C va (C

D"' hi o t (C) ' cua am so y h' N = - -X+ 1 ' d ' h-2, d 2 1 ~ nh t h d' "' A 1 va uong t ang : y = x- cat au 9-1 a1 1em x-

va B, khi do do dai do9-n AB b~ng?

Phuang phap 1: Bang bien thien

+ L?p phuong trlnh hoanh d9 giao die'm d9-ng F ( x, m) = 0 (phuong trlnh an X tham so m) + Co I?p m dua phuong trlnh ve d9-ng m = f (x)

+ L?p bang bien thien cho ham so y = f (X)

+ Dt.ra va gia thiet va bang bien thien ill do suy ra m

Phuang phap 2: Do thj ham sff

+ Co l?p m ho~c dua ve ham h~ng la duemg thclng vuong goc voi tr1,1c Oy + Tu do thi ham so tlm C\fC d9-i, C\fC tieu cua ham so (neu co)

+ Dt.ra vao so iao diem cua hai do thi ham so ta tlm duqc gia tri cua m thea yeu cau cua bai toan Cho ham so y = f ( x) co do thi la duong cong trong hlnh y

ve ben du6i Tlm so nghi~m cua phuong trlnh f (X+ 2020) = 2

4

D~t t = X+ 2020 => f ( t) = 2 va voi m6i nghi~m t se cho duy nhat mot nghi~m X Do do so nghi~m CUa phuong trinh f (X + 2020) = 2 cling la SO nghi~m CUa phuong trinh f ( t) = 2 va la SO giao diem CUa do thi ham SO y = f (X) VOi y = 2

Quan sat do thi ta thay co 2 giao die'm V?y phuong trlnh da cho co 2 Cho ham so y = f ( x) co bang bien thien nhu sau:

va B, khi do do dai do9-n AB b~ng?

Phuang phap 1: Bang bien thien

+ L?p phuong trlnh hoanh d9 giao die'm d9-ng F ( x, m) = 0 (phuong trlnh an X tham so m) + Co I?p m dua phuong trlnh ve d9-ng m = f (x)

+ L?p bang bien thien cho ham so y = f (X)

+ Dt.ra va gia thiet va bang bien thien ill do suy ra m

Phuang phap 2: Do thj ham sff

+ Co l?p m ho~c dua ve ham h~ng la duemg thclng vuong goc voi tr1,1c Oy + Tu do thi ham so tlm C\fC d9-i, C\fC tieu cua ham so (neu co)

+ Dt.ra vao so iao diem cua hai do thi ham so ta tlm duqc gia tri cua m thea yeu cau cua bai toan Cho ham so y = f ( x) co do thi la duong cong trong hlnh y

ve ben du6i Tlm so nghi~m cua phuong trlnh f (X+ 2020) = 2

4

D~t t = X+ 2020 => f ( t) = 2 va voi m6i nghi~m t se cho duy nhat mot nghi~m X Do do so nghi~m CUa phuong trinh f (X + 2020) = 2 cling la SO nghi~m CUa phuong trinh f ( t) = 2 va la SO giao diem CUa do thi ham SO y = f (X) VOi y = 2

Quan sat do thi ta thay co 2 giao die'm V?y phuong trlnh da cho co 2 Cho ham so y = f ( x) co bang bien thien nhu sau:

Ta c6 21£ ( x )I-3 = 0 ( *) <==> 1£ ( x )I = %I nen so nghi~m phuang trlnh ( *) la so giao diem cua duemg

th~ng y = % va do thi ham SO y = 1£ (X )j Ta c6 bang bien thien ham so y = 1£ (X )jla

Khi d6 duemg th~ng y = % c~t do thi ham so y = j£ ( x )j h;ti 4 diem

Phuong phap 1: Nham nghi~m ket hqp v6i cac phuang phap gic\i toan tam thuc b~c 2

Phuong phap 2: Cl,l'C tri (khi bai toan khong co l~p duqc m va cling khong nham duqc nghi~ml xem them

C6 bao nhieu gia tri nguyen cua tham so m de' do thi cua ham so y =x3

::f::.O

Cac gia tri nguyen clia m tho a man yeu cau b~ti toan la: 0 I 11 2

Phuong phap: Ta se dua bai toan ve cac d0ng 31 chu y di'eu ki~n Cho ham so y = 2

x 1

Ta c6 21£ ( x )I-3 = 0 ( *) <==> 1£ ( x )I = %I nen so nghi~m phuang trlnh ( *) la so giao diem cua duemg

th~ng y = % va do thi ham SO y = 1£ (X )j Ta c6 bang bien thien ham so y = 1£ (X )jla

Khi d6 duemg th~ng y = % c~t do thi ham so y = j£ ( x )j h;ti 4 diem

Phuong phap 1: Nham nghi~m ket hqp v6i cac phuang phap gic\i toan tam thuc b~c 2

Phuong phap 2: Cl,l'C tri (khi bai toan khong co l~p duqc m va cling khong nham duqc nghi~ml xem them

C6 bao nhieu gia tri nguyen cua tham so m de' do thi cua ham so y =x3

::f::.O

Cac gia tri nguyen clia m tho a man yeu cau b~ti toan la: 0 I 11 2

Phuong phap: Ta se dua bai toan ve cac d0ng 31 chu y di'eu ki~n Cho ham so y = 2

x 1

Ta c6: AB = 4 <=> AB2

= 16 <=> (xB -xAt +(yB -yAt = 16 <=> 2(xB -xA)2 = 16 <=> (xB -xA)2 = 8 Suy ra: (xB +xA)2 -4xAxB = 8 =>{3-m)2 -4(1-m) = 8 <::> m2

-2m-3= 0 <::> [m = -1m=3 TH1: m = -1 Suy ra (*) tr6 thrum x2

-4x+2 = 0 ~ x = ±.J2 +2 (thoa man) TH2: m = 3 Suy ra ( *) tr6 thrum x2

- 2 = 0 ~ x = ±.J2 (thoa man) Phuong phap 1: Nham nghi~m, su dvng tam thuc bac 2

Phuong ha 2: D~t an hv ket hqp v6i cac huan hap giai toan tam thuc bac 2

So gia tri nguyen cua tham so m de duemg th~ng y = -1 c~t do thi ham so y = x4

t' -(3m + 2) t+ 3m+ 1 = 0 <=:> ( t -1) ( t -3m -1) = 0 <=:> [: : ~m +

De' duemg th~ng y = -1 c~t do thi ham so tc~ti bon diem pharr bi~t c6 horum d9 nho han 2 thl di'eu

1 0<3m+1<4 <m<1

ki~n la ~ 3 vay khong c6 gia tri nguyen m thoa man 3m+1:;t:1

0 m:;t: eho ham so y = x4

, ( t :?: 0)

Phuong tdnh thrum e +2(m-2)t+4 = 0, (1)

(em) c~t Ox tiili bon diem pharr bi~t khi ( 1) c6 hai nghi~m duang phan bi~t

~ ~~: ~ ~ lm' ~~~ > 0 <=:> j[:: ~ <=:> m < 0 S> 0 -2m+4>0 m < 2

=> T = ( -oo; 0) vay c6 9 gia tri nguyen m: thoa man

Ta c6: AB = 4 <=> AB2

= 16 <=> (xB -xAt +(yB -yAt = 16 <=> 2(xB -xA)2 = 16 <=> (xB -xA)2 = 8 Suy ra: (xB +xA)2 -4xAxB = 8 =>{3-m)2 -4(1-m) = 8 <::> m2

-2m-3= 0 <::> [m = -1m=3 TH1: m = -1 Suy ra (*) tr6 thrum x2

-4x+2 = 0 ~ x = ±.J2 +2 (thoa man) TH2: m = 3 Suy ra ( *) tr6 thrum x2

- 2 = 0 ~ x = ±.J2 (thoa man) Phuong phap 1: Nham nghi~m, su dvng tam thuc bac 2

Phuong ha 2: D~t an hv ket hqp v6i cac huan hap giai toan tam thuc bac 2

So gia tri nguyen cua tham so m de duemg th~ng y = -1 c~t do thi ham so y = x4

t' -(3m + 2) t+ 3m+ 1 = 0 <=:> ( t -1) ( t -3m -1) = 0 <=:> [: : ~m +

De' duemg th~ng y = -1 c~t do thi ham so tc~ti bon diem pharr bi~t c6 horum d9 nho han 2 thl di'eu

1 0<3m+1<4 <m<1

ki~n la ~ 3 vay khong c6 gia tri nguyen m thoa man 3m+1:;t:1

0 m:;t: eho ham so y = x4

, ( t :?: 0)

Phuong tdnh thrum e +2(m-2)t+4 = 0, (1)

(em) c~t Ox tiili bon diem pharr bi~t khi ( 1) c6 hai nghi~m duang phan bi~t

~ ~~: ~ ~ lm' ~~~ > 0 <=:> j[:: ~ <=:> m < 0 S> 0 -2m+4>0 m < 2

=> T = ( -oo; 0) vay c6 9 gia tri nguyen m: thoa man

3 CASIO [X = -2 Xet phuang tdnh hoanh d<) giao diem: x - 3x + 2 = 0 =~'" -7'>

x=1 V~y co hai giao diem

G M N 1' 91 , a cac g1ao 1em cua ' d'"' ' h d"' h' h' m o t l am so y = x-"' 2 va y = ' 7x-14 2 GQi I 1a trung die'm cua do9-n th~ng MN Tim hoanh d<) die'm I

7

2 x+

Xet phuang trinh hoanh d<) giao diem: x- 2 = 7x 214

- Su dt,mg SHIFT +SOLVE hoi;ic dua ve d9-ng x+

I 1, d' "' ' d h_, MN " ' 2 + 5 7 Do a trung 1em cua 09-n t ang nen ta co x1 = -

2- =-

2

0

' h 'nhh 'nhd" · d'"' 2 1 2x-1 h' d' ' A , 2 h"' " Xet p uang tn oa Q g1ao 1em mx+m+ = - - ; t u ap an va1 m= t ay vo

2x+1 ngJ:t:~~~ A Thu d~p ~n C v6i m = -1 tl-l~Y ~<) 2 ng~~_I!l v~y

Bu6c tinh Nh~p

2x-1 2mx+m+1=

2x+1

Tim nghi~m v6i m = 2

Tim nghi~m v6i m = -1

Xet phuang tdnh hoanh d<) giao diem: x - 3x + 2 = 0 =~'" -7'> x=1 V~y co hai giao diem

G M N 1' 91 , a cac g1ao 1em cua ' d'"' ' h d"' h' h' m o t l am so y = x-"' 2 va y = ' 7x-14 2 GQi I 1a trung die'm cua do9-n th~ng MN Tim hoanh d<) die'm I

7

2 x+

Xet phuang trinh hoanh d<) giao diem: x- 2 = 7x 214

- Su dt,mg SHIFT +SOLVE hoi;ic dua ve d9-ng x+

I 1, d' "' ' d h_, MN " ' 2 + 5 7 Do a trung 1em cua 09-n t ang nen ta co x1 = -

2- =-

2

0

' h 'nhh 'nhd" · d'"' 2 1 2x-1 h' d' ' A , 2 h"' " Xet p uang tn oa Q g1ao 1em mx+m+ = - - ; t u ap an va1 m= t ay vo

2x+1 ngJ:t:~~~ A Thu d~p ~n C v6i m = -1 tl-l~Y ~<) 2 ng~~_I!l v~y

Bu6c tinh Nh~p

2x-1 2mx+m+1= 2x+1

Tim nghi~m v6i m = 2

Tim nghi~m v6i m = -1

T~p Xck dinh cua ham so luy thua y = xa tUy thUQC vao gia tri cua a C\1 the

• v 6i a nguyen duang, t~p xac dinh la JR

• v 6i a khong nguyen, t~p xac dinh ( 0; +oo)

·:· T~p xac dinh cua ham so luy thua y =X a la ( 0; +oo) v6i mQi a E JR y=xa ,a> 0

T~p xac d:inh: ( 0; +oo)

y' = a.xa-1

> 0 Vx > 0 lim X a = 01 lim X a = +oo,

x~o+ x~+«>

y=xa ,a< 0 T~p xac dinh: ( 0; +oo)

y'=a.xa-1

<0 Vx>O limxa =+oo, limxa =0

x~o+ x~+«>

Khong co ti~m c~n Bang bien thien

Ox la ti~m c~n ngang; Oy la ti~m c~n dling Bang bien thien

X 0

+

Do thj ham so y = xa luon di qua diem I ( 1; 1)

Khaos;:ithan1somi:i y=ax (a>O,a;t=l)

T~p Xck dinh cua ham so luy thua y = xa tUy thUQC vao gia tri cua a C\1 the

• v 6i a nguyen duang, t~p xac dinh la JR

• v 6i a khong nguyen, t~p xac dinh ( 0; +oo)

·:· T~p xac dinh cua ham so luy thua y =X a la ( 0; +oo) v6i mQi a E JR y=xa ,a> 0

T~p xac d:inh: ( 0; +oo)

y' = a.xa-1

> 0 Vx > 0 lim X a = 01 lim X a = +oo,

x~o+ x~+«>

y=xa ,a< 0 T~p xac dinh: ( 0; +oo)

y'=a.xa-1

<0 Vx>O limxa =+oo, limxa =0

x~o+ x~+«>

Khong co ti~m c~n Bang bien thien

Ox la ti~m c~n ngang; Oy la ti~m c~n dling Bang bien thien

X 0

+

Do thj ham so y = xa luon di qua diem I ( 1; 1)

Khaos;:ithan1somi:i y=ax (a>O,a;t=l)

\j

il < J I

Cho hai so duang a, b voi a -=1= 1 So a thoa man d~ng thuc a a = b duqc gQi la logarit co s6 a cua b va ki hi~u la loga b

a= loga b <=> aa = b, (a,b > O,a -=1= 1)

loga 1 = o,( 0 <a-=/= 1) logaaa =a,(O<a-=1=1)

loga ba = a.loga b,( a, b > O,a-=/= 1) a

logaP ba = p.loga b

log b -log c =log.(~ J

loga b =log b loga C c

logaP b = J3 .log a b loga b + loga c = loga (be)

1 log b = - -

\j

il < J I

Cho hai so duang a, b voi a -=1= 1 So a thoa man d~ng thuc a a = b duqc gQi la logarit co s6 a cua b va ki hi~u la loga b

a= loga b <=> aa = b, (a,b > O,a -=1= 1)

loga 1 = o,( 0 <a-=/= 1) logaaa =a,(O<a-=1=1)

loga ba = a.loga b,( a, b > O,a-=/= 1) a

logaP ba = p.loga b

log b -log c =log.(~ J

loga b =log b loga C c

logaP b = J3 .log a b loga b + loga c = loga (be)

1 log b = - -

a> I

X

O<a<l

V~y a=11;b=8=>a2-b3 =-391

Phuong phap 3 D~t an ph1,1 + Lo9-i 1: m.a2

f(x) +n.af(x) +p = 0 me +nt+p=O + Lo9-i 2: f ( ag(x)) t=ag(xl f ( t) = 0

+ Lo9-i 3: m.a2f(x) + n.( a.b y(x) + p.b2f(x) = 0~ Chia ca 2 ve phuang trlnh cho a 2f(x) ho~c b 2t(x) Diiit t =( ~ r') -+mt' +nt+p = 0

Tong tat ca cac nghi~m cua phuang trlnh 9x - 4.3x + 3 = 0 la

Ta c6: gx - 4.3x + 3 = 0 ¢:> [3

x = 1 ¢:> [X = 0 3x =3 X =1 V~y ton tat ca cac n hi~m cua phuon trlnh la 1

Tinh tong T tat ca cac nghi~m clia phuang trlnh 4.9x -13.6x +9.4x = 0

V~y a=11;b=8=>a2-b3 =-391

Phuong phap 3 D~t an ph1,1 + Lo9-i 1: m.a2

f(x) +n.af(x) +p = 0 me +nt+p=O + Lo9-i 2: f ( ag(x)) t=ag(xl f ( t) = 0

+ Lo9-i 3: m.a2f(x) + n.( a.b y(x) + p.b2f(x) = 0~ Chia ca 2 ve phuang trlnh cho a 2f(x) ho~c b 2t(x) Diiit t =( ~ r') -+mt' +nt+p = 0

Tong tat ca cac nghi~m cua phuang trlnh 9x - 4.3x + 3 = 0 la

Ta c6: gx - 4.3x + 3 = 0 ¢:> [3

x = 1 ¢:> [X = 0 3x =3 X =1 V~y ton tat ca cac n hi~m cua phuon trlnh la 1

Tinh tong T tat ca cac nghi~m clia phuang trlnh 4.9x -13.6x +9.4x = 0

6x +3 2x 6x +(3+m)2x +m = 0 ¢::> • = -m

2x +1 D;;tt v f( ) X = 6x +3.2x , [o 1]

2x + 1 VOl X E ;

, 1

( 6x ln6+3.2x ln2 )( 2x + 1 )-( 6x +3.2x )( 2x ln2) Taco f ( x) =

2

6x2x (ln6 -ln2 )+ 6x ln6 + 3.2x ln2 [ J

-~ -'' -2 - - - -> 0 , V'x E 0; 1 ( 2x + 1)

Suy ra f ( x) dong bien tren [ 0; 1 J iir d6 suy ra yeu diu bEd toan tuang duang v6i 2::::; -m::::; 4

6x +3 2x 6x +(3+m)2x +m = 0 ¢::> • = -m

2x +1 D;;tt v f( ) X = 6x +3.2x , [o 1]

2x + 1 VOl X E ;

, 1

( 6x ln6+3.2x ln2 )( 2x + 1 )-( 6x +3.2x )( 2x ln2) Taco f ( x) =

2

6x2x (ln6 -ln2 )+ 6x ln6 + 3.2x ln2 [ J

-~ -'' -2 - - - -> 0 , V'x E 0; 1 ( 2x + 1)

Suy ra f ( x) dong bien tren [ 0; 1 J iir d6 suy ra yeu diu bEd toan tuang duang v6i 2::::; -m::::; 4

Phuong phap 1 Dua ve d911g phuong trlnh logarit co ban log a x = b <=> x =a b (a =1= 1, a, x > 0)

{0 <a =1= 1 loga f(x) = loga g(x) <=> f{x) = g(x) > 0

DK: 0<x<5; log6 x(s-x) =1 <=> x -5x+6=0 <=> x=3

Di'eu ki~n: x > 3

log2 ( x- 3) + log2 ( x -1) = 3 <=> log2 ( x- 3 )( x -1) = 3

dfeu ki~n suy ra phuong trlnh c6 nghi~m x = 5

Tong tat ca cac nghi~m cua phuong trlnh log2 ( 3.2x -1) = 2x + 1 bfug 1

Phuong phap 2 D~t an phv: f (log a g ( x)) = 0 t=log g(x) ) f ( t) = 0 Tong cac nghi~m cua phuong trlnh log12

DK: 0<x<5; log6 x(s-x) =1 <=> x -5x+6=0 <=> x=3

Di'eu ki~n: x > 3

log2 ( x- 3) + log2 ( x -1) = 3 <=> log2 ( x- 3 )( x -1) = 3

dfeu ki~n suy ra phuong trlnh c6 nghi~m x = 5

Tong tat ca cac nghi~m cua phuong trlnh log2 ( 3.2x -1) = 2x + 1 bfug 1

Phuong phap 2 D~t an phv: f (log a g ( x)) = 0 t=log g(x) ) f ( t) = 0 Tong cac nghi~m cua phuong trlnh log12

Nh?n thay x = lla 1 nghi~m Nen phuang trlnh 2x = 2 -log3 x co nghi~m duy nhat la x 1

Bat phuong trlnh mil ca b2m co d9ng ax> b (ho~c ax~ b1ax < b1ax::;, b) voi a> 01a :;t: 1 Ta xet bat phuang trlnh co d<;lng a X > b ( 1)

+ Neu b::;, 0 I t?p nghi~m cua bat phuang trlnh la 1R

a> 1 x >lo b b > 0 thi ( 1) ¢:::>ax >a log, b ¢:::> ga

{0 <a< 1 X< loga b T~p nghi~m cua bat phuang trlnh

Nh?n thay x = lla 1 nghi~m Nen phuang trlnh 2x = 2 -log3 x co nghi~m duy nhat la x 1

Bat phuong trlnh mil ca b2m co d9ng ax> b (ho~c ax~ b1ax < b1ax::;, b) voi a> 01a :;t: 1 Ta xet bat phuang trlnh co d<;lng a X > b ( 1)

+ Neu b::;, 0 I t?p nghi~m cua bat phuang trlnh la 1R

a> 1 x >lo b b > 0 thi ( 1) ¢:::>ax >a log, b ¢:::> ga

{0 <a< 1 X< loga b T~p nghi~m cua bat phuang trlnh

T~p nghi~m cua bat phuong trinh 2x +4.5x -4 <lOX la

2x +4.5x -4 <lOx ¢:> 2x -lOx +4.5x -4 < 0 ¢:> 2x (1-5x )-4(1-5x) < 0

¢>(1-5x)(2x -4)<0

1- 5x < 0 {5x > 1 2x -4 > 0 2x > 4 [X > 2

¢> { ¢> { <=> ¢>xe(-oo;O)u(2;+oo) 1- 5x > 0 5x < 1 X < 0

Ket hqp v6i di'eu ki~n m < 5 => -% ~ m < 5 V~y c6 6 gia tri nguyen m th6a man

Bat phuang trlnh logarit ca ban c6 de:mg loga x > b (ho~c loga x 2 b,loga x < b,loga x ~b) v6i a> O,a =t: 1

Bat phuang trlnh log a x > b <=> {

a >1 x>ab

{0 <a <1 0 < x <ab

So nghi~m nguyen thUQC ( -20; 20) cua bat phuang trlnh log2 ( log4 X) 2log4 (Iog2 X)

T~p nghi~m cua bat phuong trinh 2x +4.5x -4 <lOX la

2x +4.5x -4 <lOx ¢:> 2x -lOx +4.5x -4 < 0 ¢:> 2x (1-5x )-4(1-5x) < 0

¢>(1-5x)(2x -4)<0

1- 5x < 0 {5x > 1 2x -4 > 0 2x > 4 [X > 2

¢> { ¢> { <=> ¢>xe(-oo;O)u(2;+oo) 1- 5x > 0 5x < 1 X < 0

Ket hqp v6i di'eu ki~n m < 5 => -% ~ m < 5 V~y c6 6 gia tri nguyen m th6a man

Bat phuang trlnh logarit ca ban c6 de:mg loga x > b (ho~c loga x 2 b,loga x < b,loga x ~b) v6i a> O,a =t: 1

Bat phuang trlnh log a x > b <=> {

a >1 x>ab

{0 <a <1 0 < x <ab

So nghi~m nguyen thUQC ( -20; 20) cua bat phuang trlnh log2 ( log4 X) 2log4 (Iog2 X)

¢::> {: > 1

log0,5 ( log2 ( 2x -1)) > 0 ( x > 1) <=> log2 ( 2x -1) > 1 <=> 2x -1 > 2 <=> x > %

Bat phuang trinh 3log8 ( x + 1) -log2 ( 2-x) ~ 1 c6 t~p nghi~m S = [a; b) Tinh

Ta c6: 3log8 ( x + 1) -log2 (2- x) ~ 1 <=> log2 ( x + 1) ~ 1 + log2 (2- x)

X + 1 ~ 2 ( 2 - X) {X ~ 1 1 ¢::> ¢::> ¢::> :::;; X < 2

2-x>O x<2 Khi d6 a=1, b=2

V~y P=2a2 -ab+b2

¢::> {: > 1

log0,5 ( log2 ( 2x -1)) > 0 ( x > 1) <=> log2 ( 2x -1) > 1 <=> 2x -1 > 2 <=> x > %

Bat phuang trinh 3log8 ( x + 1) -log2 ( 2-x) ~ 1 c6 t~p nghi~m S = [a; b) Tinh

Ta c6: 3log8 ( x + 1) -log2 (2- x) ~ 1 <=> log2 ( x + 1) ~ 1 + log2 (2- x)

X + 1 ~ 2 ( 2 - X) {X ~ 1 1 ¢::> ¢::> ¢::> :::;; X < 2

2-x>O x<2 Khi d6 a=1, b=2

V~y P=2a2 -ab+b2

3(log2 ( x+3)-1) s 3(log2 ( x+ 7)-log2 (2-x)) <=> log2 ( x + 3) -1 s log2 ( x + 7) -log2 ( 2- x) <=> log2 ( x + 3) + log2 ( 2- x) s log2 ( x + 7) + 1 <=> (x+3)(2-x) s 2(x+7)

V6i f( t) = e +t; f'(t) = 2t+1 > 0 v6i t E [2;+oo) nenham dong bien tren t E [2;+oo) Nen Min£( t) = £(2) = 6

Do d6 de bat phuong trlnh log2 ( sx -1) log2 ( 2.5x -2) 2 m c6 nghi~m v6i mQi x 2 1 thl:

Sn =A(1+rf

M6i thang gtri dung cling m9t so tien vao m9t thai gian co dinh Cong thuc tinh: Dau m6i thang gili vao ngan hang so tien A dong v6i lai kep r% /thang thl so tien nh~n duqc ca von I3.n lai sau n thang ( n E N * ) la sn

sn = ~[(1+rf -1](1+r)

Cong thuc tinh: Glri ngan hang so tien la A dong v6i lai suat r% /thang M6i thang vao ngay

3(log2 ( x+3)-1) s 3(log2 ( x+ 7)-log2 (2-x)) <=> log2 ( x + 3) -1 s log2 ( x + 7) -log2 ( 2- x) <=> log2 ( x + 3) + log2 ( 2- x) s log2 ( x + 7) + 1 <=> (x+3)(2-x) s 2(x+7)

V6i f( t) = e +t; f'(t) = 2t+1 > 0 v6i t E [2;+oo) nenham dong bien tren t E [2;+oo) Nen Min£( t) = £(2) = 6

Do d6 de bat phuong trlnh log2 ( sx -1) log2 ( 2.5x -2) 2 m c6 nghi~m v6i mQi x 2 1 thl:

Sn =A(1+rf

M6i thang gtri dung cling m9t so tien vao m9t thai gian co dinh Cong thuc tinh: Dau m6i thang gili vao ngan hang so tien A dong v6i lai kep r% /thang thl so tien nh~n duqc ca von I3.n lai sau n thang ( n E N * ) la sn

sn = ~[(1+rf -1](1+r)

Cong thuc tinh: Glri ngan hang so tien la A dong v6i lai suat r% /thang M6i thang vao ngay

Sn =A(1+r) -X r

Vay ngan hang so ti'en la A dong voi HH suat r%/thang Sau dling m<)t thang ke tlr ngay vay, b~t dau hoan nQ'; hai Ian hoan nQ' each nhau dung ffiQt thang, m6i Ian hoan nQ' so tien la X dong va tra het tien TIQ' sau dung n thang

Cong thuc tinh: Cach tinh so ti'en con l<;ti sau n thang giong hoan toan cong thuc tinh g{ti ngan (1+rf -1

hang va rut tien hang thang nen taco Sn =A ( 1 + r f -X r (l+rf -1 De' sau dung n thang tra het nQ' thi sn = 0 nen A(1+rf -X = 0

r A(1+rf r

Cong thuc tinh: Cach tinh so ti'en con l<;ti sau n thang giong hoan toan cong thuc tinh g{ti ngan (1+rf -1

hang va rut tien hang thang nen taco Sn =A ( 1 + r f -X r (l+rf -1 De' sau dung n thang tra het nQ' thi sn = 0 nen A(1+rf -X = 0

r A(1+rf r

(1+rf -1

3 Jxadx= xa+l+C(a:t=-1) a+1

f 1 1

4 -dx= +C

ax = - + C

Ina xdx=sinx+C

kIna

21 J cos(ax+ b )dx =~sin( ax+ b )+C 22 J sin (ax + b) dx = -~cos (ax + b)+ C

I L:5 ltantta:)(+bJc1x = _.!_lnlcos(ax+ b)l+c a

24 J cot(ax + b)dx =~!nisin (ax+ b)l + c

25 J ( ) dx = -tan ax+ b)+ C cos2

26 J ( ) dx = cot( ax+ b +C s1n2 ax+ b a

128 f( 1 + cot' (ax +b)) dx = -~co t (ax + b)+ C

• Buoc 1: ChQn X= <p( t) I trong d6 <p( t) la ham so rna ta chQn thich hqp (xem 6 bang dau hi~u)

• Buoc 2: Lay vi phan hai ve: dx = <pI ( t) dt

• Buoc 3: Bien doi f(x)dx theo t,dt gia su £( x )dx = g( t )dt

• Buoc 4: Khi d6 tinh: J f(x)dx = J g(t)dt = G(t) +C

3 Jxadx= xa+l+C(a:t=-1) a+1

f 1 1

4 -dx= +C

ax = - + C

Ina xdx=sinx+C

kIna

21 J cos(ax+ b )dx =~sin( ax+ b )+C 22 J sin (ax + b) dx = -~cos (ax + b)+ C

I L:5 ltantta:)(+bJc1x = _.!_lnlcos(ax+ b)l+c a

24 J cot(ax + b)dx =~!nisin (ax+ b)l + c

25 J ( ) dx = -tan ax+ b)+ C cos2

26 J ( ) dx = cot( ax+ b +C s1n2 ax+ b a

128 f( 1 + cot' (ax +b)) dx = -~co t (ax + b)+ C

• Buoc 1: ChQn X= <p( t) I trong d6 <p( t) la ham so rna ta chQn thich hqp (xem 6 bang dau hi~u)

• Buoc 2: Lay vi phan hai ve: dx = <pI ( t) dt

• Buoc 3: Bien doi f(x)dx theo t,dt gia su £( x )dx = g( t )dt

• Buoc 4: Khi d6 tinh: J f(x)dx = J g(t)dt = G(t) +C

Phuong phap chung:

D~t x = lalsint; vOi t E [ -~ ;~] ho~c x = laicost; VOi t E [O;n]

Di;it x=~;voi tE[-!!:;!!:J\{0} hoi;ic x = J i

Phuong phap chung:

D~t x = lalsint; vOi t E [ -~ ;~] ho~c x = laicost; VOi t E [O;n]

Di;it x=~;voi tE[-!!:;!!:J\{0} hoi;ic x = J i

1J5-t 7 1f( 7 8)

=>l=-6 -3-.t dt=-18 5t -t dt s(s-3x2r (s-3x2r

Phuong phap chung:

• Buoc 1: Ta bien doi tich phan ban dau ve dc;tng l = f f (X) dx = f ~ (X) iz (X) dx • Buoc 2: Dat: {u = ~ (x) ~{du =t;(x)dx

· d v = f2 ( x) dx v = J f2 ( x) dx • Buoc 3: Khi do : f u.dv = u.v-f v.du

Cach d?t u trong phuang phap tich phan tUng phan:

HQ nguyen ham cua ham so f (X) =X In 2x la

Phuong phap chung:

• Buoc 1: Ta bien doi tich phan ban dau ve dc;tng l = f f (X) dx = f ~ (X) iz (X) dx • Buoc 2: Dat: {u = ~ (x) ~{du =t;(x)dx

· d v = f2 ( x) dx v = J f2 ( x) dx • Buoc 3: Khi do : f u.dv = u.v-f v.du

Cach d?t u trong phuang phap tich phan tUng phan:

HQ nguyen ham cua ham so f (X) =X In 2x la

2sin2x