phân loại và phương pháp giải đại số 10

NGUYEN ANH TRUdNG - NGUYEN PHU KHANH DAU THANH KY - NGUYEN MINH NHIEN NGUYEN TAN SIENG - DO NGOC THUY (Nhdm gi£o viSn chuy§n todn THPT) PHAN LOAI & PH V0N6 PHAP 6IAI 10 1/' (Tai han On nb^t c6 cbinh If vk ho sung) Danh cho hoc sinh Idp 10 on tap va nang cao kien thufc Bien soan theo noi dung, chifofng trinh sach gi^o khoa Bo GD & DT Cac em hoc sinh than men! 'Than loai va phuong phap giai Dai so 10" la mot trong nhung cuon thuoc bo sach "Phdn loai vd phuong phdp gidi: lap 10, 11, 12 ", do nhom tac gia chuyen toan THPT bien soan. V6i each viet khoa hoc va sinh dong giup ban doc tiep can v6i mon Toan mpt each tu nhien, khong ap luc, ban doc tro nen tu tin va nang dong hon; hieu ro ban chat, biet each phan tich de tim ra trong tam cua van de va biet giai thich, lap luan cho tung bai toan. Sir da dang ciia he thong bai tap va tinh Huong giup ban doc iuon hiVng thu khi giai toan. Tac gia chii trong bien soan nhCing cau hoi mo, no! dung co ban bam sat sach giao khoa va cau true de thi dai hoe, dong thoi phan bai tap thanh cac dang toan co loi giai chi tiet. Hien nay de thi dai hoc khong kho, to hop cua nhieu van de don gian, nhung chiia nhieu cau hoi mo ne'u khong nam chae ly thuyet se lung tiing trong viee tim loi giai bai toan. Vol mot bai toan, khong nen thoa man ngay voi mot loi giai minh vua tim duoc ma phai co gang tim nhieu each giai nha't cho bai toan do, moi mot each giai se co them phan kien thuc moi on tap. Mon Toan la mot mon rat ua phong each tai tu, nhung phai la tai tu mot each sang tao va thong minh. Khi giai mot bai toan, thay vi diing thoi gian de luc Ipi tri nho, thi ta can phai suy nghl phan tich de tim ra phuong phap giai quyet bai toan do. Doi voi Toan hoc, khong co trang sach nao la thua. Tung trang^ tung dong deu phai hieu. Mon Toan doi hoi phai kien nhan va ben bi ngay tu nhCrng bai tap don gian nha't, nhirng kien thuc ca ban nha't. Vi chinh nh&ng kien thuc co ban moi giup ban doc hieu duoc nhi>ng kien thuc nang cao sau nay. Ludwig Van Beethoven: "Giot nuoc co the lam mon tang da, khong phai vi giot nude co sue manh, ma do nuoc ehay lien tuc ngay dem. Chi co su phan da'u khong met moi moi dem lai tai nang. Do do ta eo the khSng dinh, khong nhich tung bude thi khong bao gid cd the'di xa ngan dam". Mae dii tac gia da danh nhieu tam huye't cho cudn sach, song su sai sdt la dieu khd tranh khdi. Chung toi rat mong nhan duoc su phan bien va gdp y quy bau eiia quy doe gia de nhu-ng Ian tai ban sau eud'n sach duoc hoan thien hon. Thay mat nhdm bien soan Nguyen Phu Khanh. Cty TNHH MTV DWH Khang Vi.'( l^huon^ 1 MENH DE - TAP HOOP §1. MENH DE VA MENH DE CHUA BIEN A. TOM TATLY THUYET <i)inhnghia: ^ r;,^ ; ,! ; ' Mln/j (fe la mot eau khang dinh Diin^hoac Sfli. ,, ,-, ^ Mot menh de khong the vua diing hoae vua sai ^ r 2. M$nh de phu dinh: Cho menh de P. Menh de "Khong phai P " goi la menh dephu dinh cua P. Ky hieu la P. Ne'u P dung thi P sai, ne'u P sai thi P dung 3. Menh de keo theo vd menh de ddo Cho hai menh de P va Q. NJenh de "ne'u P thi Q" goi la menh dekeo theo Ky hieu la P => Q. Menh de P => Q chi sai khi P dung Q sai Cho menh de P => Q. Khi dd menh de Q => P goi la menh de ddo eua P=*Q 4. Menh de tuorng duong Cho hai menh de P va Q. Menh de "P ne'u va chi ne'u Q" goi la menh de tuong duang Ky hieu la Pc^Q. ;.S M^nh de P <=> Q dung khi ca P => Q va Q => P ciing diing Chu y: "Tuong duong" con duge goi bang cac thuat ngu khae nhu "dieu kien can va dii", "khi va chi khi", "ne'u va chi ne'u". 5. Menh de chiia bien Menh de ehiia bien la mot cau khMng djnh chua bien nhan gia trj trong mpt tap X nao dd ma vdi moi gia tn cua bien thuoc X ta duoc mgt m^nh de. Vi dy: P(n): "n chia het cho 5" vdi n la sd'tu nhien P{x; y):" 2x + y = 5" vdi X, y la so thuc 6. Cac ki hi^u V , 3 menh de phu dinh cua menh de cd chua ki hieu V ,3 . Ki hieu V: doc la vdi moi, 3: doc la ton tai Phii djnh cua m^nhde"VxeX,P(x) " la menh de " 3x e X,P(x)" Phu djnh ciia menh de"3x€X,P(x) " la menh de " Vx e X,P(x) " ^ , Phdn loai va phuamg phdpgiai Dai so'10 B. CAC DANG TOAN VA PHl/QNG PHAP GIAI. |DANG TOAN l. XAC DINH MENH DE VA TJNHDUNG SAI CIJA MENHDE Ca 1. CAC Vf DU MINH HOA Vi d\ 1: Cac cau sau day, cau nao la menh de, cau nao khong phai la menh de? Ne'u la menh de hay cho biet menh de do diing hay sai. (1) O day dep qua! (2) Phuong trinh - 3x +1 = 0 v6 nghiem (3) 16 khong la so nguyen to (4; .ai phuong trinh - 4x + 3 = 0 va x^ - Vx + 3 + 1 = 0 c6 nghiem chung. (5) So 7t CO Ion hon 3 hay khong? (6) Italia v6 dich Worldcup 2006 (7) Hai tam giac bang nhau khi va chi khi chiing c6 dien tich bang nhau. (8) Mot tu giac la hinh thoi khi va chi khi no c6 hai duong cheo vuong goc v6i nhau. Lai gidi Cau (1) va (5) khong la menh de(vi la cau cam than, cau hoi) Cac cau (3), (4), (6), (8) la nhung menh de dung Cau (2) va (7) la nhifng menh de sai. Vi du 2: Cho ba menh de sau, voi n la so tu nhien (1) n + 8 la so chinh phuong (2) Chu so tan ciing ciia n la 4 (3) n-l la so chinh phuong Biet rang c6 hai menh de diing va mot menh de sai. Hay xac djnh menh de nao dung, menh de nao sai. Lcn gidi Ta CO so chinh phuong c6 cac dm so tan cung laO, 1, 4, 5, 6, 9 . Vi vay - Nhan thay giua menh de (1) va (2) c6 mau thuan. Boi vi, gia sir 2 menh de nay dong thai la dung thi n + 8 c6 chir so tan ciing la 2 nen khong the la so chinh phuong. Vay trong hai menh de nay phai c6 mot mf nh de la diing va mpt m^nh de la sai. - Tuong tu, nhan thay giiia menh de (2) va (3) cung c6 mau thuan. Boi vi, gia sir m?nh de nay dong thoi la diing thi n - 1 c6 ch\x so tan cung la 3 nen khong the la so chinh phuong. Vay trong ba menh de tren thi menh de (1) va (3) la diing, con menh de (2) la sai. IT Cty TNHH Ml v t>\ Kliang \ ga 2. BAI TAP LUY|N TAP Bai 1.0: Cac cau sau day, cau nao la menh de, cau nao khong phai la m^nh de? Ne'u la menh de hay cho biet menh de do dung hay sai. a) Khong di\xgc di loi nay! . , ? '• . b) Bay gio la may gio? ;n:|r'c<Ho y c) Chieh tranh the gioi Ian thii hai ket thiic nam 1946.Ir „:; v! =\ d) 16chia3dul. e) 2003 khong la so nguyen to. ,. £) v'5 la so v6 ti. g) Hai duong tron phan biet c6 nhieu nhat la hai diem chung. ^ Huang dan gidi Cau khong phai m^nh de la a), b) d,^ - Cau d), f) la menh de dung. Cau e) sai. Cau g) diing Bai 1.1: Tai Tiger Cup 98 c6 bon doi lot vao vong ban ket: Vi^t Nam, Singapor, Thai Lan va Indonexia. Truoc khi thi dau vong ban ket, ba ban Dung, Quang, Trung du doan nhu sau: Dung: Singapor nhi, con Thai Lan ba. Quang: Viet Nam nhi, con Thai Lan tu. . Trung: Singapor nhat va Indonexia nhi. Ket qua, moi ban du doan dung mot doi va sai mpt doi. Hoi moi dpi da dat giai may? Huang dan gidi ,, - Ta xet du doan ciia ban Dung + Ne'u Singgapor nhi thi Singapor nhat la sai do do Indonexia nhi la dung(mau thuan) + Nhu vay Thai lan thii ba la dung suy ra Vi?t Nam nhi, Singapor nhat va Indonexia thii tu DANG TOAN 2: CAC PHEP TOAN VE MENH DE. Cac phep todn menh deduce sit dung nham muc dich ket not cac menh delai vox nhau tao ra mot menh de moi. Mot so cac phep todn menh de la : Menh de phu dinhiphep phu djnh), menh dekeo theoiphep keo theo), menh deddo, menh detitang duongiphep tuvngducmg). Phan lo^i vA phucmg phdp gidi Dai sd'lO Ca 1. CAC Vf DU MINH HOA Vi 1: Neu m?nh de phii djnh cua cac m^nh de sau, cho bie't mf nh de nay diing hay sai? P: " Hinh thoi c6 hai duong cheo vuong goc vai nhau" Q: " 6 la so nguyen to" R: " Tong hai canh cua mpt tarn giac Ion han canh con lai" S: "5>-3" K: "Phuong trinh x'* - 2x^ + 2 = 0 c6 nghi^m " H: " (x/3 - x/l2)^ - 3 " Lai gidi Ta CO cac m^nh de phii dinh la P: "Hai duong cheo ciia hinh thoi khong vuong goc voi nhau", menh de nay sai Q : "6 khong phai la so'nguyen to", menh de nay dung R: "Tong hai canh ciia mot tarn giac nho han hoac bang canh con lai", mf nh de nay sai S: " 5 < -3", menh de nay sai K: " phuang trinh - 2x^ + 2 = 0 v6 nghi^m", m|nh de nay dung vi x'^-2x^ +2 = (x^ if +l>0 H: " (V3-Vi2) =3 ", minhde nay sai Vi dv 2: Phat bieu m^nh de P => Q va phat bieu m^nh de dao, xet tinh diing sai ciia no. a) P: "Tu giac ABCD la hinh thoi" va Q:" Tu giac ABCD, AC va BD cat nhau tai trung diem moi duong" b) P: "2>9" va Q:"4<3" c) P: "Tam giac ABC vuong can tai A" va Q:" Tam giac ABC c6 A = 2B " d) P: "Ngay 2 thang 9 la ngay Quoc Khanh ciia nuac Viet Nam" va Q: "Ngay 27 thang 7 la ngay thuong binh lift si" Lai gidi a) Menh de P Q la "Neu tu giac ABCD la hinh thoi thi AC va BD cat nhau tai trung diem moi duang", menh de nay dung. M?nh de dao la Q ri> P: "Neu tu giac ABCD c6 AC va BD c^t nhau tai trung diem moi duang thi ABCD la hinh thoi", menh de nay sai. b) Menh de P => Q la "Neu 2 > 9 thi 4 < 3", menh de nay diing vi menh de P sai. Menh de dao la Q => P: "Neu 4 < 3 thi 2 > 9", menh de nay diing vi m^nh de c) Menh de P => Q la "Neu tam giac ABC vuong can tai A thi A = 2B ", menh de nay diing Menh de dao la Q => P: "Neu tam giac ABC c6 A = 2B thi no vuong can tai A", menh de nay sai d) Menh de P => Q la "Neu ngay 2 thang 9 la ngay Quoc Khanh ciia nuoc Viet Nam thi ngay 27 thang 7 la ngay thuong binh liet sT" ' ' Menh de dao la Q => P: "Neu ngay 27 thang 7 la ngay thuong binh liet sT thi ngay 2 thang 9 la ngay Quoc Khanh ciia nuoc Viet Nam" Hai menh de tren deu diing vi menh de P, Q deu diing Vi du 3: Phat bieu menh de P o Q bang hai each va xet ti'nh diing sai ciia no a) P: "Tii giac ABCD la hinh thoi" va Q: "Tu giac ABCD la hinh binh hanh CO hai duong cheo vuong goc voi nhau" b) P: "Ba't phuong trinh Vx^ -3x > 1 c6 nghiem" va Q: " ^(-1)^ -3.(-l) > 1 " Lai gidi a) Ta CO menh de P <=> Q diing vi menh de P => Q, Q => P deu diing va duoc phat bieu bang hai each nhu sau: "Tii giac ABCD la hinh thoi khi va chi khi tii giac ABCD la hinh binh hanh CO hai duong cheo vuong goc voi nhau" va "Tu giac ABCD la hinh thoi neu va chi neu tii giac ABCD la hinh binh hanh CO hai duang cheo vuong goc voi nhau" b) Ta CO menh de P <=> Q diing vi menh de P, Q deu dung(do do m^nh de P => Q, Q => P deu diing) va dugc phat bieu bSng hai each nhu sau: "Bat phuong trinh \/x^ -3x > 1 c6 nghiem khi va chi khi -3.(-\) > 1" va "Bat phuong trinh -3x >1 c6 nghiem neu va chi neu ^(-1)^ -3.(-l) > 1" ea 2. BAI TAP LUYIN TAP Bai 1.2: Neu menh de phu dinh ciia cac menh de sau, cho bie't menh de nay diing hay sai? P: "Trong tam giac tong ba goc bang 180"" Q:"(V3-V27)^ la so nguyen" R: "Viet Nam v6 djch Worldcup 2020" ' ' Phdn loai va phuong phdpgidi Dai so 10 S:"-^>-2" 2 K: "Bat phuong trinh x^°" > 2030 v6 nghiem " Huong dan gidi ' '^Mat* Ta CO cac menh de phu dinh la P: " Trong tam giac tong ba goc khong bang 180°", menh de nay sai Q : " (N/3 - \/27) khong phai la so nguyen ", menh de nay sai R:" Viet Nam khong v6 djch Worldcup 2020", menh de nay chua xac djnh dirge diing hay sai S: "<-2", menh de nay dung K :" Bat phuong trinh x^"^'^ > 2030 c6 nghiem ", menh de nay dung Bai 1.3: Phat bieu menh de P => Q va phat bieu menh de dao, xet tinh diing sai ciia no. a) P: "Tu giac ABCD la hinh chii nhat" va Q: "Tu giac ABCD c6 hai duong thang AC va BD vuong goc v6i nhau" b) P: "-V3>-V2" vaQ: "{-^fsf > {-^f " c) P: "Tam giac ABC c6 A = B + C " va Q: "Tam giac ABC c6 BC^ = AB^ + AC^ " d) P: "To Hiru la nha Toan hoc Ion cua Viet Nam" va Q: "Evariste Galois la nhk Tha loi lac cua The'gioi" Huang dan gidi a) P => Q: " Neu tu giac ABCD la hinh chu nhat thi tu giac ABCD c6 hai duong thang AC va BD vuong goc voi nhau", menh de sai Q => P: " Neu tu giac ABCD hai duong thang AC va BD vuong goc vol nhau thi tu giac ABCD c6 la hinh chCr nhat", m?nh de sai b) P ^ Q:" Neu -73 > -y/2 thi [-^f > [-^f ", menh de dung P ^ Q:" Neu (-^f > (-^flf thi - > -72 ", menh de sai c) P=>Q: "Neu tam giac ABC c6 A = B + C thi tam giac ABC c6 BC?=AB?+AC?" Q => P: "Neu tam giac ABC c6 BC^ = AB^ + AC^ thi A = B + C " Ca hai menh de deu diing. d) P^Q:" Neu To Huu la nha Toan hoc Ion cua Viet Nam thi Evariste Galois la nha Tho loi lac cua The'gioi", Q => P :" Neu Evariste Galois la nha Tho loi lac ciia The'gioi thi To Hiru la nha Toan hpc Ion ciia Viet Nam ". Hai m^nh de diing. Cty TNHHMTV nvv)l Khang Vict Bai 1.4: Phat bieu m^nh de P <=> Q bang hai each va va xet tinh diing sai ciia no a) Cho tu giac ABDC. Xet hai menh de | P: "Tu giac ABCD la hinh vuong". Q: "Tii giac ABCD la hinh chu nhat c6 hai duong cheo vuong goc voi nhau". b) P: "Bat phuong trinh x^ -3x + l >0 c6 nghiem" va Q: " Bat phuong trinh x^ - 3x +1 < 0 v6 nghiem" Huang dan gidi a) Ta CO menh de P <=> Q diing vi menh de P Q, Q => P deu diing va dugc phat bieu bang hai each nhu sau: "Tu giac ABCD la hinh vuong khi va chi khi tii giac ABCD la hinh chu nhat CO hai duong eheo bang vuong goc voi nhau " va "Tu giac ABCD la hinh vuong neu va chi neu tu giac ABCD la hinh chii nhat CO hai duong cheo vuong goc voi nhau " b) Ta CO menh de P o Q sai vi menh de P diing con Q sai. Phat bieu menh de P <=> Q b3ng hai each "Bat phuong trinh x^ - 3x +1 > 0 c6 nghiem khi va chi khi ba't phuong trinh x^ - 3x +1 < 0 v6 nghiem" va "Bat phuong trinh x^ - 3x +1 > 0 c6 nghiem neu va chi neu ba't phuong trinh x^ - 3x +1 < 0 v6 nghiem" Bai 1.5: Cho cac menh de : < . a-\/3 A : "Neu AABC deu c6 canh bang a, duong cao la h thi h = " ; B : "Tii giac CO bon canh bSng nhau la hinh vuong" ; C: "15 la so nguyen to" ; D : " >/T25 la mpt so nguyen". a) Hay cho bie't trong cac menh de sau, menh de nao diing, menh de nao sai : A =^ B, A ^ D, B =^ C. b) Hay cho bie't trong cac menh de sau, mf nh de nao diing, menh de nao sai : A<=> B, B C, B o D. Huang dan gidi Ta CO A va D la cac menh de diing, B va C la cac m^nh de sai. Do do ; a) Menh de A=>B sai vi A diing va B sai. ' > Menh de A D diing vi A va D deu diing. . , Menh de B =:> C diing vi B sai. b) Menh de A B sai vi menh de A B sai (Hoac A diing va B sai). Menh de B <=> C diing vi hai menh de B va C deu sai. Menh de A Ci> D diing vi hai m^nh de A va D deu diing. Phiht loai va pltucnig phiip giiii Dai so'10 Bai 1.6: Hay phat bieu menh de keo theo P => Q, Q => P va xet tinh diing sai cua menh de nay- a) Cho tiV giac ABCD va hai menh de: ' ' • P: "Tong 2 goc doi cua tu giac loi bang 180" " va Q: " Tu giac noi tiep duoc duong tron ". b) P :" >/2 - > -1" va Q: "(^/2 - Vsf > (-1)' " Htt&ttg dan gidi a) P => Q : " Neu tong 2 goc doi cua tiV giac loi bang 180" thi tiV giac do npi tiep du(?c duong tron ". Q P: "Neu tu giac khong noi tiep duong tron thi tong 2 goc doi cua tu giac do bang 180"" Menh de P => Q dung, menh de Q => P sai. b) P Q: " Neu V2 - V3 > -1 thi (V2 - Sf > {-if " Q=>P:"Neu (V2 - Vs)^ < (-1)^ thi V2->/3>-l " Menh de P => Q sai vi P dung, Q sai, menh de Q => P diing vi P va Q deu dung. DANG TOAN 3: MENH DE CHUA BIEN VA MENH DE CHUA KiHIEU V, 3. ffll.CAC Vl DU MINH HOA Vi du 1: Cho m^nh de chua bien "P(x): x > x"'", xet tinh dung sai ciia cac menh de sau: a) PO) c) VxeN, P(x) b) P d) 3xeN, P(x) Lai gidi a) Ta CO P(l): 1 > 1^ day la menh de sai b) Taco p''' 1 3J 1 : — > 3 day la menh de diing c) Ta CO Vx € N, X > x'"* la m^nh de sai vi P(l) la menh de sai d) Ta CO 3x € N, x > x'' la mfnh de diing vi x - x"' = x(l - x)(l + x) < 0 voi moi so' ty nhien. Cty TNHH MTV DWH Khang Viet Vi du 2: Dung cac ki hieu de vie't cac cau sau va vie't menh de phu djnh ciia no. a) Tich ciia ba so'tu nhien lien tiep chia he't cho sau b) Voi mpi so thuc binh phuong ciia la mot so khong am. i c) Co mot so nguyen ma binh phuong ciia no bang chi'nh no. d) Co mot so hCru ti ma nghich dao ciia no Ion hon chinh no. Loi gidi i-uij \ a) Ta CO P: Vn e N, n(n + l)(n + 2):6, menh de phii djnh la r P:3neN, n(n + l)(n + 2)/6. ^'/^ b) Ta CO Q : Vx e M, x^ > 0, menh de phii djnh la Q :3x 6 K, x^ < 0 c) Ta CO R : 3n G Z, n^ = n , menh de phii djnh ia R : Vn e Z, n^ ;t n .'' ' d) 3q € Q, — > q , menh de phu dinh la Vq e Q, i < q . q M Vi du 3: Xac djnh tinh diing sai ciia menh de sau va tim phii dinh ciia no : a) A: "VxeR, x^ > 0 " b) B: " Ton tai so tu nhien deu la so nguyen to". c) C: " 3x6N, X chia he't cho x + 1 " d) D:"Vn£N, n'^-n^+l la hop so" e) E: "Ton tai hinh thang la hinh vuong ". f) F: "Ton tai so thuc a sao cho a +1 + —< 2" a + 1 Lai gidi a) Menh de A diing va A : 3x € R, x' < 0 b) Menh de B diing va B : "Voi moi sotu nhien deu khong phai la so nguyen to" c) Menh deC sai va C: " Vx e N,x/(x + 1)" d) Menh de D sai vi voi n = 2 ta c6 n'* - n^ +1 = 13 khong phai la hop so M^nh de phii djnh la D: " 3n e N, n"* - n^ +1 la so so nguyen to" e) Menh de E diing va E : " Voi mpi hinh thang deu khong la hinh vuong ". 0 Menh de F diing va menh de phii djnh la F: "Voi moi so thuc a thi a + l + >2 . a + 1 Phdtt loai va phucnig phdp gM Dai so 10 Ca 2. BAI TAP LUYfN TAP Bai 1.7: Xet diing (sai) menh de va phu dinh cac menh de sau : a) VxeM, -x^ +1 >0 b) Vx € K, x"* - x^ + 1 = (x^ + 73x + l)(x^ - N/3X +1) c) 3x e N, n +3 chia het cho 4 d) 3qeQ, 2q2-l=0 e) 3nGN,n(n + l) la mpt so chinh phuong Huang dan gidi a) Menh de Vx G M, x'' - x^ +1 > 0 sai ch^ng han khi x = -1 ta c6 (-1)'-(-!)'+1 = -1<0. M^nh de phu dinh la 3x e R, x'' - x^ +1 < 0 . b) Menh de Vx e x"" - x^ +1 = (x^ + V2x + l)(x^ - N/2X +1) dung vi x'' - x^ +1 = (x^ + ])^ - 3x^ = (x^ + x/3x + l)(x2 - >/3x + l) Menh de phu djnh la 3x e IR, x"* - x^ +1 9^ (x^ + VSx + l)(x^ - VSx +1). c) Menh de 3x G N, n^ + 3 chia het cho 4 diing vi n = 1 e N va n^ + 3 = 4:4 M^nh de phii dinh la " Vx e N, n + 3 khong chia het cho 4" d) Men de 3q e Q, 2q^ -1 = 0 sai. Menh de phu dinh la Vq e Q, 2q^ -1 0 e) Menh de "3nGN,n(n + l) la mot so chinh phuong" diing. M|nh de phii dinh la " Vn e N, n (n +1) khong phai la mot so chinh phuong" Bai 1.8: a) Voi neN, cho menh de chua bien P(n):"n^ +2 chia het cho 4". Xet tinh diing sai ciia menh de P(2007). b) Xet tinh diing sai cua menh de P{n) : " 3n e N*, ^n(n +1) chia het cho 11". ^ Huang dan gidi a) Ta CO : Voi n = 2007 thi n^ + 2 = 2007^ + 2 la so le nen khong chia het cho 4. Vay P(2007) la menh de sai. b) Xet bieu thiVc ^ , voi n G N*. ta c6 : 2 Voi n = 10 thi = 55 : chia het cho 11. Vay menh de da cho la menh de diing. | Cty TNHH MTV DWH Khang Vjc-i Bai 1.9: a) Cho menh de P : "Voi moi so thuc x, neu x la so hiiu ti thi 2x la sd'hiru ti". Dung ki hieu viet P, P va xac djnh tinh diing - sai ciia no. b) Phat bieu MD dao ciia P va chiing to MD do la diing. Phat bieu MD duoi dang MD tuong duong Huang dan gidi 'f i ' sr a) Menh deP"VxG R,XGQ=>2X GQ" .MD diing. P: "3x G R,x 6 Q 2x g Q". MD sai b) MD dao ciia P la " Voi moi so thuc x, XGQ khi va chi khi 2XGQ". Hay " Vx e R,x G Q <:> 2x G Q". 'O riff' Bai 1.10: Cho sotu nhien n. Xet hai menh de chiia bien: •'• 'i' A(n): "n la sochan", B(n) : "n^ la so ch§n". a) Hay phat bieu menh de A(n) B(n). Cho biet menh de nay diing hay sai ? b) Hay phat bieu m?nh de " Vn G N, B(n) A(n) ". c) Hay phat bieu m^nh de "Vn e N, A(n) <r> B(n)". Huang dan gidi a) A(n) => B(n) : "Neu n la so chan thi n^ la so chan". Day la menh de diing, vi khi do n = 2k (k G N) => n^ = 4k2 la so chan. b) " Vn G N, B(n) => A(n) " : Voi moi so tu nhien n, neu n^ la so chan thi n la so chan. c) " Vn € N, A(n) <=> B(n) " : Voi moi so tu nhien n, n la so chin khi va chi khi n^ la sochan. Bai 1,11: Xet tinh diing sai ciia cac m^nh de sau: a) P:"V\GR,VyGR:x + y = l" b) Q :"3x G R,3y G R : x + y = 2" c) R:"T GR,VyGR:x + y=3" d) S :" Vx G R,3y G R : x + y = 4" Huang dan gidi ' ' a) Menh de P sai vi ch^ng han X = 1GIR, y = 2GK nhung x + y ^ 1 b) Menh de Q diing vi X = y = 1 => x + y = 2 •> c) Vi X + y = 3 nen voi moi y G M thi luon ton tai x = 3 - y do do mf nh de R diing ' d) Menh de S diing " Phatt loai va fhucntg phdp giai D(it so §2: AP DUNG MENH DE VAO SUY LUAN TOAN HOC A. TOM TAT LY THUYET /. 't)inh li va. chung minh dinh U. • Trong toan hoc djnh ly la mot menh de diing. Nhieu dinh ly dugc phat bieu duoidang "VxeX, P(x)=:>Q(x)", P(X),Q(X) la cac menh de chiia bien • Co hai each de chiing minh dinh li duoi dang tren Cach 1: Chung minh true tie'p gom cac budc sau: - Lay xeXba'tkyma P(x) dung - Chung minh Q(x) dung(bang suy luan va kien thuc toan hoc da biet) Cach 2: Chung minh bang phan djnh li gom cac buoc sau: - Gia su ton tai XQ e X sao cho P(x„) dung va Q(X(,) sai - DiJng suy luan va cac kien thuc toan hoc de di de'n mau thuan. 2. Djnh li dao, dieu ki?n can, dieu ki?n dii, dieu ki^n can va du. • Cho djnh li duoi dang "Vx 6 X, P(x) =:> Q(x)" (1). Khi do P(x) la dieti kien dii deed Q(x) Q(x) \a dieu kien can de CO P(x) • Menh de Vx€X, Q(x)=> P(x) dung thi duoc goi dinh li dao ciia djnh li dang (1) Liic do (1) duoc goi la dinh Ixj thuan va khi do c6 the gop lai thanh mot djnh li Vx G X, Q(x) o P(x), ta goi la " P(x) la dieu kien can va dii de c6 Q(x)" Ngoai ra con noi" P(x) ne'u va chi neu Q(x)"," P(x) khi va chi khi Q(x)", B. CAC DANG TOAN VA PHl/QNG PHAP GIAI. DANG TOAN 1: PHUONG PHAP CHUNG MINH BANG PHAN CHUNG. QLCAC V( DUMINH HOA Vi du 1: Chung minh rSng voi moi so tu nhien n, n'' chia het cho 3 thi n chia he't cho 3. Lai giai Gia su n khong chia he't cho 3 khi do n = 3k +1 hoac n = 3k + 2, keZ ; Voi n = 3k + 1 ta c6 n^ = (3k + if = 27k^ + 27k^ + 9k +1 khong chia he't cho 3 (mau thuan) Voi n = 3k + 2 ta c6 n^ =(3k + 2)^ =27k^+54k^+36k + 4 khong chia het cho ba (mau thuan) Vay n chia he't cho 3. 1L Chf TNHH MTV DVVH Khang Vi^t Vi du 2: Cho tarn thuc f (x) = ax^ + bx + c, a 0 . Chung minh rang ne'u ton tai so thuc a sao cho a.f (a) < 0 thi phuong trinh f (x) = 0 luon c6 nghiem. Ta CO f(x) = a X + • 2a Lai gidi ^-,A = b2-4ac. 4a Gia su phuong trinh da cho v6 nghiem, nghTa la A < 0. Khi do ta co: af(x) = a' ^ b^2 X + • 2a \ itiv.tn >0, VxeR 4 *,»vwi' Suy ra khong ton tai a de af (a) < 0 , trai voi gia thiet. Vay dieu ta gia su 6 tren la sai, hay phuong trinh da cho luon c6 nghiem. Vi du 3: Chung minh rang mot tarn giac c6 duong trung tuye'n vua la phan giac xua't phat tu mot dinh la tam giac can tai dinh do. Lai gidi Gia su tam giac ABC co AH vua la duong trung tuyen vua la duong phan giac va khong can tai A. Khong mat tinh tong quat xem nhu AC > AB . Tren AC lay D sao cho AB = AD . Goi L la giao diem cua BD va AH . Khi do AB = AD, BAL = LAD va AL chung nen AABL = AADL Do do AL = LD hay L la trung diem cua BD B Suy ra LH la duong trung binh cua tam giac CBD => LH//DC dieu nay mau thuan vi LH,DC Ccit nhau tai A Vay tam giac ABC can tai A. ; Q 2. BAI TAP LUYEN TAP Bai 1.12: Chung minh b^lng phuong phap phan chung: Ne'u phuong trinh bac hai ax^ + bx + c = 0 v6 nghiem thi a va c cung da'u. Huong dan gidi Gia su phuong trinh v6 nghiem va a, c trai da'u. Voi dieu ki^n a, c trai da'u CO a.c<0 suy ra A = b^ - 4ac = b^ + 4(-ac) > 0 Nen phuong trinh co hai nghiem phan biet, dieu nay mau thuan voi gia thiet phuong trinh vo nghiem. Vay phuong trinh v6 nghiem thi a, c phai cung da'u. Bai 1.13: Chung minh bang phuong phap phan chung: Ne'u hai so'nguyen duong CO tong binh phuong chia he't cho 3 thi ca hai so do phai chia he't cho 3. Phdn lo^i va phuang phdp gi&i Dai so 10 Huang dan gidi Gia su trong hai so' nguyen duang a va b c6 it nhat mot so khong chia het cho 3, chang han a khong chia het cho 3. The thi a c6 dang: a = 3k + 1 hoac a = 3k+2. Luc do a^ = 3m + 1 , nen neu b chia het cho 3 hoac b khong chia het cho 3 thi a^ + b^ cung co dang: 3n + 1 hoac 3n + 2, tuc la a^ + b^ khong chia het cho 3, trai gia thie't. Vay neu a^ + b^ chia het cho 3 thi ca a va b deu chia het cho 3. Bai 1.14: Chung minh rang : Neu do dai cac canh cua tam giac thoa man bat ding thuc a^ + b^ > 5c^ thi c la do dai canh nho nhat cua tam giac. Huong dan gidi Gia su c khong phai la canh nho nhat cua tam giac. Khong mat tinh tong quat, gia sua < c => a^ < c^ (1) Theo bat ding thuc trong tam giac, ta co b < a + c b^ < (a + c)^ (2). Doa<c^(a + c)^ <4c2 (3) Tu (2) va (3) suy ra b^ < 4c^ (4). Cong ve voi ve'(l) va (4) ta co a^ + b^ < 5c^ mau thuan voi gia thie't Vay c la canh nho nhat cua tam giac. Bai 1.15; Cho a, b, c duong nho hon I. Chung minh rang it nhat mot trong ba ba't ding thuc sau sai a(l -b) > i, b(l - c) > c(l - a) >i Huong dan gidi Gia sir ca ba bat ding thuc deu dung. Khi do, nhan theo ve'ciia cac bat ding thuc tren ta duoc: 4; 3 1 a(l-b).b(l-c).c(l-a)> - hay a(l-a).b(l - b).c(l-c) > — (*) 64 Mat khac a(l-a) = -a^+a = ^- 1 1 a — N2 1 < — 4 DoO<a<l=>0<a(l-a)<^ Tuong tu thi 0 < b(l - b) < i, 0 < c(l - c) < 1 Nhan theo ve ta dugc a(l-a).b(l - b).c(l - c) < — (**) 64 Bat ding thuc (**) mau thuan (*) Vay CO ft nhat mot trong cac bat dang thuc da cho la sai. (dpcm) Bai 1.16: Neu a]a2>2(b5+ b2) thi it nhat mot trong hai phuang trinh + ajx + bj = 0, + a2X + b2 = 0 co nghiem. Huong dan gia i Gia su ca hai phuang trinh tren v6 nghiem Khi do D, = ai^ - 4bi < 0, D] = - 4b2 < 0 _^a/-4bi+a2^-4b2 <Oc:>a,^+32^ <4(bi+b2) (1) ' Ma (a, - a2 f > 0 <=> a? + a^ > 23^32 (2) Tu (1) va (2) suy ra 23^32 < 4(bj + b2) hay 3^32 < 2{b] + bj) trai gia thie't. Vay phai co it nhat 1 trong hai so Ai, A2 Ion han 0 do do it nhst 1 trong 2 phuang trinh x + ajx + b^ =0, x + a2X + b2 = 0 co nghiem. Bai 1.17: Chung minh rang sfl la so v6 ti. Huong dan gidi ,^ De dang chung minh dugc neu n^ lasochSnthi n laso'chan Gia sir V2 la so hCru ti, tuc la J2= — , trong do m, n e N*, (m, n) = 1 n Tit N/2 = — => m^ = 2n^ => la so chan n => m la so chan => m = 2k, k e N Tir m^ = 2n^ =:> 4k^ = 2n^ :=> - 2k^ => n^ la so chin => n la so chan Do do m chan, n chan, mau thuan vai (m, n) = 1 . Vay N/2 la so v6 ti. a + b + oO (1) Bai 1.18: Cho cac so a, b, c thoa cac dieu ki§n : • ab + be + ca > 0 (2) abc > 0 (3) Chirng minh rang ca ba so a, b, c deu duong. Huong dan gidi Gia sir ba so a, b, c khong dong thai la so duang. Vay co it nhat mot so khong duang. Do a, b, c CO vai tro binh ding nen ta co the gia sir: a < 0 Neu a = 0 thi mau thuan voi (3) Ne'u a < 0 thi tir (3) => be < 0 Ta CO (2) <=> a(b + c) > -be => 3(b ^ b + c<0 a + b + c<0 mauth|ulli%y»P TiWH BiNH THUAfJ Vav ca b3 so a, b, c deu duone. Than loai va phumtg phapgiai Dai so 10 Bai 1.19: Chung minh bang phan chiing dinh li sau : "Neu tarn giac ABC c6 cac duang phan giac trong BE, CF bang nhau, thi tarn giac ABC can". Huang dan gidi • Neu B > C thi ta dung hinh binh hanh BEDF nhu hinh ve . Ta CO : B > C => B^ > Q =i> > Q (1) Ngoai ra, BE = CF => DF = CE => Di +D2 =C2 +C3 (2). Tu (1) va (2) suy ra < Q => EC < ED => EC < FB . Xet cac tarn giac BCE va CBF, ta thay : ^ BC Chung, BE = CF, BF > CE nen Q > B^ =^ C > B. Mau thuan. • Truong hop C > B, chung minh hoan toan tuong tu nhu tren. Do do B = C. Vay tarn giac ABC can tai A. Bai 1.20: Cho 7 doan thSng c6 do dai Ion hon 10 va nho hon 100. Chung minh rang luon tim dugc 3 doan de c6 the ghep thanh mgt tarn giac. Huang dan gidi Truoc het sap xep cac doan da cho theo thu tu tang dan cua do dai aj, ,87 va chung minh rang trong day da xep luon tim dugc 3 doan lien tiep sao cho tong cua 2 doan dau Ion hon doan cuoi (vi dieu kien de 3 doan c6 the ghep thanh mgt tarn giac la tong cua 2 doan Ion hon doan thu ba). Gia su dieu can chung minh la khong xay ra, nghia la dong thoi xay ra cac bat dang thuc sau: aj <a3; a2 +33 <a^; ; +a(,<ay Tu gia thie't a^, aj c6 gia tri Ian hon 10, ta nhan dugc 83 > 20 . Tu 82 > 10 va 83 > 20 ta nhan dugc a^ > 30, 35 > 50, ag > 80 va ay > 130 . Dieu ay > 130 la mau thuan voi gia thiet cac do dai nho hon 100. Co mau thuan nay la do gia sir dieu can chung minh khong xay ra. Vay, luon ton tai 3 doan lien tiep sao cho tong cua 2 doan dau Ion hon doan cuoi. Hay noi each khac la 3 doan nay c6 the ghep thanh mgt tam giac. DANG TOAN 2: SUDUNG THUAT NGU DIEU KIEN CAN, BIEU KIEN DU, DIEU KIEN CAN VADU. || Ca 1. CAC Vf DU MINH HOA " * '' ' Vi du 1: Cho djnh li: "Cho so tu nhien n. Neu n'^ chia het cho 5 thi n chia het cho 5". Dinh li nay dugc viet duoi dang P => Q. a) Hay xac dinh cac menh de P va Q. b) Phat bieu dinh li tren bang each dung thuat ngii "dieu kien can". c) Phat bieu djnh li tren b^ng each diing thuat ngii "dieu kien du". d) Hay phat bieu djnh li dao (neu c6) cua djnh li tren roi dung cac thuat ngii "dieu kien can va du" phat bieu gop ca hai dinh li thuan va dao. Lai gidi. a) P : "n la so tu nhien va n^ chia het cho 5", Q : "n chia het cho 5". b) Voi n la so tu nhien, n chia het cho 5 la dieu kien can de n'^ chia het cho 5 ; hoac phat bieu each khac : Voi n la so tu nhien, dieu kien can de chia het cho 5 la n chia het cho 5. c) Voi n la so tu nhien, n"^ chia het cho 5 la diiFu kien du de n chia het cho 5. d) Dinh li dao : "Cho so tu nhien n, neu n chia het cho 5 thi n"^ chia het cho 5". That vay, neu n = 5k thi n'^ = 5\k^: So nay chia het cho 5. Dieu kien can va du de n chia he't cho 5 la n'^ chia het cho 5. Vi du 2: Phat bieu cac menh de sau voi thuat ngii "Dieu kien can", "Dieu kien du" » f' a) Neu hai tam giac bang nhau thi chung c6 dien tich bang nhau b) Neu so'nguyen duong chia het cho 6 thi chia het cho 3 c) Neu hinh thang c6 hai duong cheo bang nhau thi no la hinh thang can d) Neu tam giac ABC vuong tai A va AH la duong cao thi AB^ = BC.BH Lai gidi a) Hai tam giac bang nhau la dieu kien du de chiing c6 dien tich bang nhau Hai tam giac c6 dien tich bSng nhau la dieu kien can de chiing bang nhau b) So'nguyen duong chia het cho 6 la dieu kien dii de no chia het cho 3 So nguyen duong chia het cho 3 la dieu kien can de no chia het cho 6 c) Hinh thang c6 hai duong cheo bang nhau la dieu ki|n du de no la hinh thang can Hinh thang can la dieu kien can de no c6 hai duong cheo bang nhau d) Tam giac ABC vuong tai A va AH la duong cao la dieu ki?n dii de AB^-BC.BH ^.R [...]... n ( T ) - n ( V n A ) - n ( A n T ) sinh chi bie't da cau la 25 -15 = 10 - n ( T n V) + n ( V n A n T ) So hoc sinh chi bie't danh cau long la 8 + 10 + 1 2 - 3 - 4 - 5 + 2 = 20 30-15 = 15 Vay nhom do c6 20 em Do do ta CO sT so'hoc sinh cua lap lOAi la Bai 1.30: Co 40 hoc sinh gioi, moi em gioi it nha't mot mon Co 22 em gioi Van, 10 + 15 + 15 = 40 V i du 2: Trong lop IOC c6 45 hoc sinh trong do c6 25... \/2015^ chinh xac den hang chuc va hang tram biet V2015^ = 25450,71 l/oc l u g n g sai so tuyet d o i t r o n g m o i truang hgp ea 1 cAc v f D u MINH HOA Vi hang d o la hang phan tram b) Ta CO 100 < 101 < 100 0 nen hang cao nhat m a d n h o h o n m o t d o n v i ciia 6 JCi hieu khoa hoc cua mot so TUONG gidi D o d o ta phai q u y t r o n so a = 2,235 den hang phan t r a m suy ra a * 2,24 K h i d o... chac chan /j.:, - N e u so'gan d i i n g la so nguyen t h i dang chuan ciia no la A .10* ^ t r o n g d o A la so nguyen, k la hang thap nha't c6 c h u sochSc ( k e N ) (suy ra m o i c h i i so cua A deu la c h u so'chic chan) ,y hang d o la hang n g h i n MQ\O thap phan khac 0 deu vie't dugc d u d i dang a l O " , 1 < |a| < 10, n e Z ) dang n h u vay dugc goi la kihieu khoa HQC cua so do DANG TOAN 1: TINH... mon: Anh, Toan, Van Co 8 em gioi Van, 10 em gioi Anh, 12 em gioi Toan, 3 em gioi Van va Toan, 4 em gioi Toan va Anh, 5 em gioi Van va Anh , 2 em gioi ca ba mon Hoi nhom do c6 bao nhieu em? Huong dan gidi Ky hi?u A la tap hop nhiing hoc sinh gioi Anh, T la tap hop nhCrng hoc sinh gioi toan, V la tap hop nhijng hoc sinh gioi Van Theogiathie'ttac6:n(V) = 8, n ( A ) = 10, n(T) = 12, n ( V n T ) = 3, n (... SO GAN HOC CUA MOT SO - ^ ^ [ d y l ^ ' r " so chac va vie't dang chuan ciia so gan diing a biet a) So nguoi dan tinh Nghe A n la a = 3214056 nguoi voi do chinh xac d = 100 nguoi g _ 1,3462 sai so tuong doi ciia a bang 1 % y\^ = 50 < 100 < r " r # = 500 nen chu so hang tram(so 0) khong la so chac, con chii so hang nghin(so 4) la chii so chac 17700 (hay vie't a * 17700) b) Ta CO 0,01 < 0,056 < 0,1... = 212m ± 2m Huang dan gidi = 43 + u, y = 63 + AN Lai gidi hang tram Nen ta phai quy tron so 17638 den hang tram Vay so quy tron la X CHU KHOA CUA MOT P I C A C ViDUMINH HOA Huang dan gidi a) Vi 10 < 16 < 100 nen hang cao nha't ma d nho han mot don vj cua hang do la Gia sir CAC CHU SO CHAC V Taco P = 2x + 2y = 2(43 + 63) + 2 u + 2 v = 212 + 2(u + v ) Theo gia thie't - 0 , 5 < u < 0 , 5 v a - 0 , 5... sang la 300 n g h i n km/s nen t r o n g v o n g m o t n a m no d i dupe p : " D i e m M nSm tren phan giac cua goc O x y " ; ; -i;-:::; 31536000.300 = 9,4608 .10' ^ k m ca 2 BAI T A P L U Y E N Huang TAP Bai 1.45: So dan cua m o t tinh la A = 103 4258 ± 300 ( n g u o i ) H a y t i m cac chi> so Hu&ng dan = 50 )'^ = ^ + (k() 2(ko+l)7t_ 3 b)An(B\A) =0 CON " xe B c) A u ( B \ A ) = A w B Lai gidi CO Vx, X G A \ < ! => x e A X«B hi' uy ra ( A \) c A • De c h u n g m i n h A . hon 10, ta nhan dugc 83 > 20 . Tu 82 > 10 va 83 > 20 ta nhan dugc a^ > 30, 35 > 50, ag > 80 va ay > 130 . Dieu ay > 130 la mau thuan voi gia thiet cac do dai nho hon 100 ta phai quy tron so a = 2,235 den hang phan tram suy ra a * 2,24 . b) Ta CO 100 < 101 < 100 0 nen hang cao nhat ma d nho hon mot don vi ciia hang do la hang nghin . do B = C. Vay tarn giac ABC can tai A. Bai 1.20: Cho 7 doan thSng c6 do dai Ion hon 10 va nho hon 100 . Chung minh rang luon tim dugc 3 doan de c6 the ghep thanh mgt tarn giac. Huang