Bồi dưỡng học sinh giỏi toán lớp 12 Môn Đại số và giải tích tập 2

ins. LE HOANH PHO Nha gido Uu tu BOIDUSNG 7 HOC SINH GIOI TOAN DAI SO - GIAI TICH • Danh cho HS lap 12 on tap & nang cao Id" nang lam bdi. Chudn bi cho cdc ki thi qudc gia do Bo GD&OT to choc. a NHA XUA1HN DAI HOC QUOC GIA HAJIOI B6I DUBNG HQC SINH QOI TOAN DAI SO-GIAI TICH Boi duOng hoc sinh gioi Toan Dai so 10-1. Boi duQng hoc sinh gioi Toan Dai so 10-2. Boi dti8ng hoc sinh gioi Toan Hinh hoc 10. Boi duQng hoc sinh gioi Toan Dai so 11. - Boi duQng hoc sinh gioi Toan Hinh hoc 11. Bo de thi tii luan Toan hoc. Phan dang va phi/dng phap giai Toan So phtic. Phan dang va phirong phap giai Toan To hop va Xac suat. 1234 Bai tap ttj luan dien hinh Dai so giai tich 1234 Bai tap tu luan dien hinh Hinh hpc luting giac ThS. LE HOANH PHO Nhd gido Uu tu BOIDUdNG , HOC SINH GIOI TOAN 3> DAI SO-GIAI TICH 12 - Ddnh cho HS ldp 12 on tdp & ndng cao ki ndng Idm bdi. Chudn bi cho cdc ki thi qudc gia do Bo GD&DT to chut:. NHA XUAT BAN DAI HQC QUOC GIA HA NQI NHA XUAT BAN DAI HOC QUOC GIA HA NOI 16 Hang Chudi - Hai Ba Triing Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896; Hanh chinh: (04) 39714899; Tong bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trach nhiem xudt bdn: Giam doc PHUNG QUOC BAO Tong bien tap PHAM THI TRAM Bien tap noi dung HAI NHU THUY NGAN Sila bdi LE HOA Che bdn CONG TI ANPHA Trinh bay bia SON KY Doi tdc lien ket xudt ban CONG TI ANPHA SACH LIEN KET BOI DUCfNG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 2 Ma so: 1L-181DH2010 In 2.000 cuon, kho 16 x 24 cm tai cong ti TNHH In Bao bi Hung Phu So'xua't ban: 89-2010/CXB/11-03/DHQGHN, ngay 15/01/2010 Quyet dinh xua't ban so: 181LK-TN/XB In xong va nop lu'u chieu quy II nam 2010. Ldi N6I DXU Be giup cho hoc sinh ldp 12 cb them tai liiu tu boi duong, ndng cao va ren luyen ki nang gidi todn theo chuong trinh phdn ban mai. Trung tdm sdch gido due ANPHA xin trdn trong giai thieu quy ban dong nghiep va cdc em hoc sinh cuon: "Boi dudng hoc sinh gidi todn Dai so' Gidi tich 12" nay. Cuon sdch nay nam trong bo sdch 6 cuon gom: - Boi duong hoc sinh gioi todn Hinh hoc 10. - Boi duong hoc sinh gidi todn Dai so' 10. - Boi duong hoc sinh gioi todn Hinh hoc 11. - Boi dudng hoc sinh gidi todn Dqi so'- Gidi tich 11. - Boi dudng hoc sinh gidi todn Hinh hoc 12. - Boi dudng hoc sinh gidi todn Gidi tich 12. do nha gido uu tu, Thac si Le Hoanh Phd to'chiec bien soqn. Noi dung sdch duoc bien soqn theo chuong trinh phdn ban: co bdn vd ndng cao mdi cua bo GD & DT, trong dd mot so"van deduac md rong vdi cdc dang bdi tap hay vd khd dephuc vu cho cdc em yeu thich mud'n ndng cao todn hoc, cd dieu kien phdt trien tot nhat khd ndng ciia minh. Cud'n sdch la sir ke thira nhirng hieu biet chuyen mdn vd kinh nghiem gidng day ciia chinh tdc gid trong qua trinh true tiep dirng ldp bdi dudng cho hoc sinh gidi cdc ldp chuyen todn. Vdi ndi dung sue tich, tdc gid da cd'gdng sap xep, chon loc cdc bdi todn tieu bieu cho tirng the loai khdc nhau ung vdi noi dung ciia SGK. Mdt sd'bai tdp cd the khd nhung cdch gidi duoc dua tren nen tdng kien thuc va ki nang co ban. Hoc sinh can tu minh hoan thien cdc ki ndng cung nhu phat trien tu duy qua viec gidi bai tap cd trong sdch trudc khi ddi chieu vdi ldi gidi cd trong sdch nay, cd the mdt soldi gidi cd trong sdch con cd dong, hoc sinh cd the'tu minh lam ro hon, chi tiet hon, ciing nhu tu minh dua ra nhirng cdch lap ludn mdi han. Cluing tdi hy vong bd sdch nay se la mdt tai lieu thie't thuc, bd' ich cho ngudi day va hoc, dqc biet cdc em hoc sinh yeu thich mdn todn vd hoc sinh chuan bi cho cdc ky thi quoc gia (tot nghiep THPT, tuye'n sinh DH - CD) do bd GD & DT to chirc sap tdi. Trong qua trinh bien soqn, cudn sdch nay khdng the tranh khdi nhung thieu sdt, chiing tdi rat mong nhdn duoc gdp y ciia ban doc gan xa debb sdch hoan thien hon trong ldn tdi ban. Moi y kien dong gop xin lien he: Trung tam sach giao due Anpha 225C Nguyen Tri Phuong, P.9, Q.5, Tp. HCM. - Cong ti sach - thiet bj giao due Anpha 50 Nguyen Van Sang, Q. Tan Phii, Tp. HCM. DT: 08. 62676463, 38547464 . Email: alphabookcenter@yahoo.com Xin chan thanh cam on! 3 MUC LUC Chuong II: Ham so luy thira ham so mu va ham so logarit §2. Phucmg trinh, he phuomg trinh, bat phuong trinh mu va logarit 5 Dang 1: Phuong trinh mu va logarit 5 Dang 2: Bat phuong trinh mu, logarit 21 Dang 3: He phuong trinh mu, logarit 31 Chuong HI: Nguyen ham, tich phan va ung dung § 1. Nguyen ham 46 Dang 1: Dinh nghia va tinh chat 47 Dang 2: Phuong phap bien doi bien so 55 Dang 3: Nguyen ham tirng phan 62 §2. Tich phan 70 Dang 1: Dinh nghia va tinh chat 71 Dang 2: Tich phan da thuc, phan thuc 81 Dang 3: Tich phan luong giac 89 Dang 4: Tich phan can thiic 100 Dang 5: Tich phan mu - logarit 108 §3. Ung dung cua tich phan 124 Dang 1: Tinh dien tich hinh phang 124 Dang 2: Tinh the tich vat thS 130 Chirong IV. So phuc §1. S6 phuc 139 Dang 1: Phep toan ve so phuc 140 Dang 2: Bieu dien va tap hop diem 143 §2. Can bac hai va phucmg trinh 151 Dang 1: Can bac hai ciia so phuc 151 Dang 2: Phuong trinh nghiem phuc 156 §3. Dang luong giac 165 Dang 1: Viet dang luong giac 165 Dang 2: Toan ung dung 171 CHUONG II: HAM SO LUY THUA, HAM SO MU VA HAM SO LOGARIT §2. PHUONG TRiNH, H£ PHUONG TRINH, BAT PHUONG TRINH MU VA LOGARIT A. KIEN THUG CO BAN Phuong phap chung: - Dua ve cung mot co so - Dat an phu - Logarit hoa, mu hoa - Su dung tinh chat cua ham so B. PHAN DANG TOAN DANG 1: PHUONG TRlNH MU VA LOGARIT , a ^ = a g(x) (a>0) log a f(x) = logag(x), (a > 0, a * 1) o - Phuong trinh mu co ban: a x = b (a > 0, a * 1) Neu b < 0, phuong trinh vo nghiem Neu b > 0, phuong trinh co nghiem duy nhat x = log a b. a = l a * 1 , f(x) = g(x) - Phuong trinh logarit co ban: log a x = b(a>0, a^l) Phuong trinh logarit co ban luon co nghiem duy nhat x = a b . f(x)>0 hayg(x)>0 f(x) = g(x) Chii y: Ngoai 4 phuong phap chinh de giai phuong trinh mu, lograrit, ta co the dung dinh nghTa, bien doi thanh phuong trinh tich so, dung do thi, bit dang thuc, Vi du 1: Giai cac phuong trinh sau: a) 2 x2 ' 3x+2 =4 b) (2 + V3 ) 2x = 2 - V3 c) 2 x+1 5 X = 200 d) 0,125.4 2x-3 = (4V2) x Giai a) PT o 2* 2 - 3x+2 =2 2 ox 2 -3x + 2 = 2»x 2 -3x = 0ox = 0 hoac x = 3. b) PT<^(2+ V3) 2x = (2+ S)' 1 <=>2x = -l ox = -i. 2 c) PT o 2 . 10 x = 200 o 10 x = 100 o x * 2 5x 5x d) PT <=> 2" 3 . 2 4x " 6 = 2 T o 2 4x " 9 = 2 Y 5x <=> 4x - 9 = — <=> 8x - 18 = 5x o x = 6. 2 -BDHSG DSGT12/2- c Vi du 2: Giai cac phuong trinh sau: a) (1,5) 5x-7 /o\ x+1 b) 7 X -' = 2 X 1 3 x+— x+— c) 9 X - 2 2 =2 ->2x-l ^ ylogx _ ^logx+1 _ ^ jlogx-1 _ J3 j^ogx-\ Giai CO 5x-7 _ CO -x-l u, ,2j O 5x - 7 -lox = l b)PT»f = 2 x 7<=> 7 <=> x = log, 7 1 i - A x+- c) PTo 9 X +i.9 x =2 X+2 +2.2 X+2 9 X = 3.2" +2 3 3 = — O X - 1 = lOgg — O X = 1 - lOgg 2 . Z 2 1 2 d) PT » 7 logx +13.7 logx = 5 logx . 5 + 3.5 logx o 7 logx 1 + 13 -logx 5 + ? . 5, 20 ^ ylogx | | _ glogx <=> logx = 2 <=> x = 100. Vi du 3: Giai cac phuong trinh sau: a) 4 X - 2 X - 6 = 0 c) e 2x -3e x -4+ 12e" x = 0 28 1_ 5 20 b) 3 X + 1 + 18.3" x = 29 d) 27 x + 12 X = 2.8 X Giai a) Dat t = 2 X , (t > 0) thi PT <=> t - t - 6 = 0 Chon nghiem t = 3 <^> 2 X =3 ox = log 2 3 1R b) Dat t = 3\ t > 0 thi PT o 3t + — = 29 <=> 3t 2 - 29t + 18 = 0 o t = 9 hoac t = - Giai ra nghiem x = 2 hoac c = log32 - 1. c) Dat t = e\ (t > 0) thi PT <=> t 2 - 3t - 4 + — =0 O t 3 - 3t 2 - 4t + 12 = 0 <=> (t - 2)(t + 2)(t - 3) = 0. Chon nghiem t = 2 hoac t = 3 nen x = ln2 hoac x = ln3. 6 -BDHSG DSGT12I2- d) Chia 2 ve cho 8 X > 0 thi PT: 27V (12\* ( 3 T i '3^ - + - {2 \ 2) - 2 = 0 . Dat t = -2 ,t>0. PT <=> t 3 + t - 2 = 0 <=> (t - l)(t 2 + t + 2) = 0o t = 1 <=> x = 0. Vi du 4: Giai cac phuong trinh: a)2.25 x + 5.4 x = 7.10 x b) 4 x + 6 x = 9 c) U/2-V3 +U2+S) =4 d) 4 X+Nx2 " 2 -5.2 x " 1+VxI:2 =6 a) PT«5||] -7 Giai 2 = 0. Dat t = ,t>0. PT <=> 5t 2 - 7t + 2 = 0 o t = 1 hoac t = | (thoa man) Suy nghiem x = 0 hoac x = 1. b) Dieu kien x * 0, dat y = — va chia hai ve cho 4 y , ta co: x PT« -1 = 0<=> 1 + 75 . 1 + 75 —-— es> y = log 3 —-— 1 , 1 + 75 1 , o — = log 3 —— c=>- = log 3 X 2 2 X 2 1 + 75 «x = log^_ i - 2 c) Ta co 72-73.72 + 73 = 1, dat t = (72 + 73") , t > PT<^t+-=4«t 2 -4t+l=0 t o t = 2 + 73 hoac t = 2- 73»x = 2 hoac x = -2. d) Dat t = 2 x+vx2 - 2 , t > 0 thi PT <=> t 2 - -1 = 6 2 <=> 2t - 5t - 12 = 0. Chon nghiem t = 4. x nen x + 7x 2 -2 = 2 <=> 7x 2 - 2 =2- <=> 2 - x > 0 va x 2 - 2 = 4 - 4x + x 2 <=> x < 2 va x <=> x = - 2 2 -BDHSG DSGT12/2- 1 Vi du 5: Giai cac phuong trinh: a) x 3 +(x-2) 6 =0 b) V2 X V4 X (0,125) X = 4^2 c) Ul6-x + $fx~+~l=3 d) 3 /xTl- 3 /x^T = v / x 2 -1 Giai a) BK:x>0, x-2>0<=>x>2. Voi x > 2 thi VT > 0 nen PT vo nghiem. b) DK: x # 0, PT o 2 2 _ 17 » 22.23"^ =2 3 »- + —= - 2 3 2x 3 , 1 » 5x - 14x - 3 = 0 <=> x = — hoac x = 3. 5 c) DK:-1 <x< 16. Datu= #L6-x, v= 7x + l thi u, v > 0. lu + v = 3 1 6 7 £ 2x— 6 7 2 2x (2~ 3 ) x = 2 3 o 2 2 2 x = 2 3 k > Ta co he: u 4 +v 4 =17 Dat S = u + v, P = uv thi u 4 + v 4 = (u 2 + v 2 ) 2 = ((u + v) 2 - 2uv) 2 - 2u 2 v 2 17 = (9-2P) 2 -2P 2 = 2P 2 -36P + 81. Do do P = 2 hoac P = 16. Vi S 2 - 4P > 0 nen chon P = 2 suy ra S = 3 nen nghiem x = 1 hoac x = 15. d) DK: x < -1 hoac x > 1. Vi x = ±1 khong la nghiem nen dieu kien: x < -1 hoac x > 1. Ta co x la nghiem thi -x cung la nghiem, do do xet x > 1. PT<» 7(x + l) 2 -7(x-l) 2 = v / x Y ^lci>6 x + 1 x-l x-l Vx+1 ep^i . t > 0 thi PT <i> t - - = 1 <=>t 2 -t-l = 0 x-l t Chon nghiem t 1 + V5 suy ra nghiem ciia PT cho la: x = ± 't + r 6 t-1 VOI t = 1 + V5 Vi du 6: Giai cac phuong trinh sau: a) 3 4 * = 4 3 " c) S"" 1 ^ 2 =8.4 X b) 3 X .8 X+1 =36 5 15 -BDHSG DSGT12/2- [...]... a) 3 2x+1 o 22 -2 4 2x+1 X - 5.6 < 0 b) 2 " " - 2 ~ X 2x2 4x 2 2x x2+1 - 2 0 Bat phuong trinh t - — - 2 < 0 ot -2t-4 — < x < - hay x > 12x-8 12x-8 6 2 3 51 Ket hop D K ta co nghiem: — < x < 12 2 2 6 e 2 8 2 2 28 -BDHSG DSGT12I2- b) D K : x > - 3... thi BPT: V8 + 2 t - t > 5 - 2t 2 "5-2t0 2 5-2t>0, 8 + 2 t - r >0 5 - 0 thi 0 < t - 2t < 3 2 -1 < t < 3 ft -2t-30 2 t < 0 hay t > 2 2 Do do 2 < l - l < 3 o log 2 . log 22 log 2 x + log 2 log 22 x = 2- o ilog 2 log 2 x + log 2 f^log 2 x j = 2 11 3 o -log 2 log 2 x + log 2 - + log 2 log 2 x = 2 <=> -log 2 log 2 x = 3. o log 2 log 2 x. phuong trinh sau: a) 2 x2 ' 3x +2 =4 b) (2 + V3 ) 2x = 2 - V3 c) 2 x+1 5 X = 20 0 d) 0 , 125 .4 2x-3 = (4V2) x Giai a) PT o 2* 2 - 3x +2 =2 2 ox 2 -3x + 2 = 2 x 2 -3x = 0ox = 0 hoac . x 2 - log 2 8 = 2 log 2 x-3 log 2 (4x) = 2 2 logj 4 + logi x V 2 2 J = ( -2- log 2 x) 2 = (2 + log 2 x) 2 Dat t = log 2 x thi PT <=> (2 + t) 2 + 2t - 3 = 8 ot 2