III. PHAN PHOI TH 6I LUONG
VTu) < gix) (2)
HI. Neu g(x) < 0 . bat phuong trinh (2) v6 nghiem, diing hay sai?
H2. Trong trudng hgp g(x) > 0 . bat phuong trinh (2) tuong duang vdi bat phuong trinh f(x) > [g(x)] dung hay sai?
GV: Ca hii cau hdi tren deu dung. Tir dd neu len each giai bat phuong trinh dang (2).
• GV neu vi du 2 va hudng din HS giii theo cac cau hdi sau day: HI. Hay tim dieu kien ciia phuang trinh.
H2. Vdi dieu kien cua bat phuang trinh thi 2x + l < 0 - phuang trinh da cho v6 nghiem, dung hay sai?
H3. Hay giai phuong trinh trong trudng hgp 2.v +1 > 0 • Thuc hien |H2
Hoat ddng cua GV Cau hdi 1
Tim dieu kien ciia phuang trinh.
Cau hdi 2
Khi X < -20, bat phuang trinh cd nghiem hay khdng?
Cau hdi 3
Khi X > - 2 0 , hay giai phuang trinh tren.
Hoat ddng cua HS Ggi y tra Idi cau hdi 1
Dieu kien cua bat phuang trinh la : x< - 2 8 - ^ 4 6 4 hoac x > - 2 8
+ V464.
Ggi y tra Idi cau hdi 2
Bat phuang trinh v6 nghiem.
Ggi y tra Idi cau hdi 3
x = 20.
• GV neu vi du 3 va hudng din HS giai theo cac cau hdi sau: HI. Tim dieu kien cua bat phuang trinh.
H2. Khi X < 2 bat phuang trinh cd nghiem hay khdng? H3. Hay giai bat phuang trinh trong trudng hgp x > 2.
Thuc hien H3
Hoat ddng ciia GV Cau hdi 1
Tim dieu kien ciia bat phuang trinh.
Cau hdi 2
Khi X < 3, bat phuang trinh cd nghiem hay khdng?
Cau hdi 3
Khi X > 3, hay giai bat phuang trinh tren.
Hoat ddng cua HS
Ggi y tra Idi cau hdi 1
Dieu kien cua bat phuang trinh la : x< - 3 hoac X > 5. (adsbygoogle = window.adsbygoogle || []).push({});
Ggi y tra Idi cau hdi 2
Bat phuang trinh vd nghiem.
Goi y tra Idi cau hdi 3
5 < x < 6 .
• GV neu vi du 4 va hudng din HS giai theo cac cau hdi sau: HI. Tim didu kien cua bat phuang trinh.
H3. Hay giai bat phuang trinh trong trudng hgp x > 3.
• Thuc hien H4
Hoat ddng ciia GV
Cau hdi 1
Tim didu kien ciia bat phuang trinh.
Cau hdi 2
Khi X < - 2 , bat phuang trinh cd nghiem hay khdng?
Cau hdi 3
Khi X > - 2 , hay giai bat phuang trinh trdn.
Hoat ddng cua HS Ggi y tra Idi cau hdi 1
Dieu kien ciia bat phuang trinh la : x< - 1 hoac X > 1.
Ggi y tra Idi cau hdi 2
Bat phuang trinh cd nghiem vdi
mgi X < - 2 .
Ggi y tra Idi cau hdi 3
S = ( - c » ; - 2 ) u [ - 2 ; - - ) = ( - 00 ;
4
- ! > •
TOM T A T BAI HOC
1. Phuong trinh ^//(x) = g(x) tuong duong vdi he:
'gix)>0
Jix) = g^ix) 2. Phuang trinh |/(x)| = ^(x) tuong duong vdi he 2. Phuang trinh |/(x)| = ^(x) tuong duong vdi he
fix)>0 gix) > 0
[/2(x) = / ( x )
3. Bat phuang trinh yjf(x) < g(x) tuong duang vdi he (adsbygoogle = window.adsbygoogle || []).push({});
/ • ( x ) > 0
gix) > 0
/{-v)<^^-v) 4. Bat phuang trinh ^ / ( x ) < ^'(.v) tuong duong vdi
{ / • ( x ) > 0
hoac gix)>0
\gix)<0 • [f(x)>g^(x)
MOT SO CAU HOI TRAC NGHIEM
Cau 1. Phuang trinh Ix - ll = 2x + 3 tuong duong vdi :
( a ) ( x - l ) 2 = ( 2 x + 3)2; (c) - ( x - l ) 2 = ( 2 x + 3)2; (b) ( x - l ) ^ = - ( 2 x + 3 ) ^ (d) ( x - l ) 2 = ( 2 x + 3)2 3 v > - - 2 Trd Idi. Chgn (d).
Cau 2. Phuang trinh v v - 1 = 2x + 3 tuong duong vdi:
( a ) ( x - l ) = (2x + 3 ) ^ (c) - ( x - l ) = (2x + 3 ) ^ (b) ( x - l ) = - ( 2 x + 3 ) ^ ,2 (d) ( x - l ) = (2x + 3)^ 3 .V > - - Trd lai. Chgn (d).
Cau 3. Bat phuang trinh Ix - ll > 2x + 3 tuang duang vdi :
(a)
x > l
( . v - l ) 2 > ( 2 x + 3 ) 2 ' (b)
X > -
(c) x > ( x - 1 ) ^ > ( 2 x + 3)^ hoac X < (d) ' Trd ( x - 1 ) ^ > ( 2 x + 3)^ 2 hoac X > X > - - I 2 Idi. Chgn (c). 3 2
Cau 4. Ba't phuang trinh |x - 1 | < 2x + 3 tuong duang vdi
3 (a) x > l ( x - 1 ) ^ <(2x + 3)2' (b) x > - ( x - l ) 2 < ( 2 x + 3)^ (c) x > ( x - 1 ) ^ <(2x + 3)2 hoac X < (d) ( x - 1 ) ^ <(2x + 3)^ x > hoac X > — 2 Trd Idi. Chgn (b).
Cau 5. Bat phuang trinh Vx - 1 > 2x + 3 tuong duong vdi:
(a) (c) x > l ( x - l ) > ( 2 x + 3) 2 ( x - l ) > ( 2 x + 3)^ 2 ' (b) 3 x > - - 2. ( x - l ) > ( 2 x + 3)^ hoac X < 148
(d)
( x - l ) > ( 2 x + 3)^
x . - l hoac X > — 2
Trd lai. Chgn (c).
Cau 6. Bat phuang trinh Vx - 1 < 2x + 3 tuong duong vdi
3 (a) x > l | ( x - l ) < ( 2 x + 3)2 ' 3 (b) x > 2 ( x - l ) < ( 2 x + 3)' (c) (d) x > ( x - l ) < ( 2 x + 3)' ( x - l ) < ( 2 x + 3)^ x > hoac X < hoac X > Trd Idi. Chgn (b).
HUdNG DAN B A I TAP SGK
Bai 65.
De gidi dang todn ndy, HS can: (adsbygoogle = window.adsbygoogle || []).push({});
- On tap lai cdch gidi phuang trinh vd bdt phuang trinh chita ddu gid tri tuyet ddi.
- On tap lai cdch gidi phuang trinh vd bdt phuang trinh chita ddu can bdc hai.
- Xem lai cdc vi du vd HD trong SGK.
Hudng ddn.
a) Dieu kien cua phuang trinh la x< - 5 hoac x > - 1 . Phuang trinh tuong duong vdi :
Dap sd. X = 11
b) Dieu kien cua phuang trinh la : x > — Phuang trinh tuang duong vdi :
Dap sd. X = — 3
( 3 x - 2 ) = 0
c) Dieu kien ciia bat phuong trinh la : x >
2 Dapsd. S = [-l ; 4 ] .
d) Binh phuang hai ve' ta cd : ( v - 1 ) (2x +1) > 0 Tap nghiem la:
Bai 66.
Degidi dang todn ndy. HS can:
- On tap lai cdch gidi phuang trinh vd bdt phuang trinh chita ddu can bdc hai.
- Xem lai cdc vi dii vd HD trong SGK.
Hudng ddn.
.- . ,, -2-V6 -2 + V6
a) Dieu kien cua phuang trinh la x< hoac x > Phuang trinh tuong duang vdi :
2
x ^ + 2 x - 2 = 0 x > - l
Dap sd. x = V3 - 1
uM^-~- , ' ' u .^ur ^ - 1 0 1 - V 9 1 7 7 ^ . ^ - 1 0 1 + V9177
b) Dieu kien cua phuang trinh la x< hoac x >
8 8
Phuang trinh tuong duong vdi:
J21x-336 = 0
| x > - 1 0
Dap sd. X = 16.
c) Xem hudng din trong SGK. Dap so. X = - 1 ± V2 (adsbygoogle = window.adsbygoogle || []).push({});
d) Xem hudng din trong SGK.
-3±V37
Dap sd. X
z Bai 67.
De gidi dang todn ndy, HS edn:
- On tap lai cdch gidi bdt phuang trinh chita ddu edn bdc hai. - Xem lai cdc vi du vd HD trong SGK.
Hudng ddn.
a) Dieu kien cua bat phuang trinh la x< - 3 hoac x > 2. Bat phuang trinh tuong duong vdi:
| 3 x - 7 < 0
X > 1
Dap sd. S = 2 ;
b) Dieu kien ciia bat phuang trinh la x > — 2 Bat phuang trinh tuong duong vdi:
4 x ^ - 1 4 x + 1 0 > 0
3 x > - x > -
hoac X >
[5 1
Dap sd. S = — ; + oc .
L2 J
c) Dieu kien cua bat phuo'ng trinh la x< Bat phuang trinh tuong duong vdi :
,, . f x ^ + 2 x - 2 > 0
X > 1 hoac <
[x<l
Dap sd. S = (-GO ; - 1 - Vs ) u ( - 1 + V3 ; + oo).
d) Dieu kien cua bat phuang trinh la x< - 2 hoac x > 7 Bat phuang trinh tuong duang vdi :
2 V2' X < — hoac 2 3x^ + X + 1 5 < 0 1 x > - Dap sd. S = (-00 ; - 2 ] . Bai 68.
De gidi dang todn ndy, HS edn:
- On tap ve mien xdc dinh ciia hdm so.
- On tap lai cdch gidi bdt phuang trinh chita ddu edn bdc hai.
- Xem lai cdc vi dii vd HD trong SGK. Hudng ddn.
a) Dieu kien cua bat phuong trinh la (adsbygoogle = window.adsbygoogle || []).push({});
Bit phuang trinh tuong duong vdi:
x^ + 3x - 4 - X + 8 > 0
x ^ + 3 x - 4 > 0
.2 hoac
x ^ + 2 x + 4 > 0 Dap sd. Tap xac dinh ciia ham sd la R.
. v ^ + 3 x - 4 < 0 - . v ^ - 4 x + 1 2 > 0
b) Dap sd. Tap xac dinh ciia ham sd la (-00; - - ) u (3 ; + co).
c) Dap sd. Tap xac dinh cua ham sd la 0 ; 7 - V 2 9 u 7 + V29 + 00