Tài liệu BỒI DƯỠNG HỌC SINH GIỎI TOÁN ĐẠI SỐ GIẢI TÍCH 12 TẬP 2

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i n s . LE H O A N H P H O Nha gido Uu tu B O I H O C D A I D U S S S I O N N G H - 7 G G I I O A I T I O T • 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 X U A 1 H N DAI HOC QUOC GIA HAJIOI A I N C H 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. L E H O A N H P H O Nhd gido Uu tu B O I D H O C U S I d N N H G , G I O I T O A N 3> D A I S O - G I 1 2 A I T I - 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 C H NHA XUAT B A N D A I 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 P H A M T H I T R A M Bien tap noi HAI NHU dung THUY NGAN Sila bdi LE HOA Che bdn CONG T I A N P H A Trinh bay bia SON K Y Doi tdc lien ket xudt ban CONG T I A N P H A 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 t i 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 I I nam 2010. L d i N 6 I D X U 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 sin 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 M U C L U C Chuong I I : Ham so luy thira ham so mu va ham so logarit §2. Phucmg trinh, he phuomg trinh, bat phuong trinh mu va logarit Dang 1: Phuong trinh mu va logarit 5 5 Dang 2: Bat phuong trinh mu, logarit 21 Dang 3: He phuong trinh mu, logarit 31 Chuong H I : 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 I V . 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 C H U O N G II: H A M S O L U Y T H U A , H A M S O M U V A H A M SO LOGARIT §2. P H U O N G T R i N H , H £ P H U O N G T R I N H , B A T PHUONG TRINH M U V A LOGARIT A. K I E N T H U G C O B A N Phuong phap chung: - Dua ve cung mot co so - Dat an phu - Logarit hoa, mu hoa - Su dung tinh chat cua ham so B. P H A N D A N G T O A N DANG 1: PHUONG TRlNH M U VA LOGARIT - Phuong trinh mu co ban: a = b (a > 0, a * 1) Neu b < 0, phuong trinh vo nghiem Neu b > 0, phuong trinh co nghiem duy nhat x = log b. a = l (a>0) , ^ = a a * 1 , f ( x ) = g(x) x a g(x) a - Phuong trinh logarit co ban: log x = b ( a > 0 , a ^ l ) Phuong trinh logarit co ban luon co nghiem duy nhat x = a . f(x)>0 hayg(x)>0 log f(x) = logag(x), (a > 0, a * 1) o f(x) = g(x) a b a 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,... V i du 1: Giai cac phuong trinh sau: a) 2 ' x 2 c) 2 a) PT o =4 b) (2 + V3 ) 5 = 200 d) 0,125.4 Giai 3 x + 2 x + 1 X 2* 2 3x+2 2 x 2x-3 = 2 - V3 = (4V2) x = 2 o x - 3 x + 2 = 2 » x - 3 x = 0 o x = 0 hoac x = 3. 2 b) P T < ^ ( 2 + V 3 ) 2 2 <=>2x = - l o x = - i . 2 c) PT o 2 . 10 = 200 o 10 = 100 o x * 2 5x 5x d) PT <=> 2" . 2 " = 2 o 2 " = 2 5x <=> 4x - 9 = — <=> 8x - 18 = 5x o x = 6. 2 -BDHSG DSGT12/2- c 2 x = (2+ S ) ' x 3 1 x 4x 6 T 4x 9 Y V i du 2: Giai cac phuong trinh sau: /o\ a) (1,5) 5x-7 x + 1 b) 7 -' = 2 X 1 3 c) 9 - 2 x+— = 2x+— X ^ ylogx _ ^logx+1 _ ^ jlogx-1 _ J3 j^ogx-\ ->2x-l 2 X CO 5x-7 u , b)PT»f =2 -x-l ,2j X x =2 - l o x = l O 5x - 7 7<=> x 1 9 +i.9 3 c) P T o _ CO Giai 7 <=> x = log, 7 i +2.2 X + 2 X+2 A -.9 3 x+= 3.2" X +2 = — O X - 1 = lOgg — O X = 1 - lOgg 2 . 2 2 Z d) PT » 7 logx 1 +13.7 logx .- = 5 logx . 5 + 3.5 logx .- o 7 28 5 ^ ylogx | 20 | _ glogx logx 1+ 13 -logx 5 ? . 5, + 1_ 20 <=> logx = 2 <=> x = 100. V i du 3: Giai cac phuong trinh sau: b) 3 + 1 + 18.3" = 29 a) 4 - 2 - 6 = 0 X c) e 2 x X X d) 2 7 + 12 = 2.8 - 3 e - 4 + 12e" = 0 x x x x X X Giai a) Dat t = 2 , (t > 0) thi PT <=> t - t - 6 = 0 X Chon nghiem t = 3 <^> 2 =3 o x = log 3 1R b) Dat t = 3 \ t > 0 thi PT o 3t + — = 29 X 2 <=> 3t - 29t + 18 = 0 o t = 9 hoac t = 2 Giai ra nghiem x = 2 hoac c = log32 - 1. c) Dat t = e\ (t > 0) thi PT <=> t - 3t - 4 + — 2 =0 O t - 3t - 4t + 12 = 0 <=> (t - 2)(t + 2)(t - 3) = 0. 3 2 Chon nghiem t = 2 hoac t = 3 nen x = ln2 hoac x = ln3. 6 -BDHSG DSGT12I2- d) Chia 2 ve cho 8 > 0 thi PT: X 27V (12\* ( T i' 3 ^ - 2 = 0 . Dat t = + {2 \ 2) 3 -2 ,t>0. PT <=> t + t - 2 = 0 <=> (t - l ) ( t + t + 2) = 0 o 3 t = 1 <=> x = 0. 2 Vi du 4: Giai cac phuong trinh: a)2.25 + 5.4 = 7.10 x x c) U/2-V3 +U2+S) b) 4 x =4 d) 4 x +6 X+Nx2 x =9 " -5.2 " 2 x 1+VxI:2 =6 Giai a) P T « 5 | | ] -7 2 = 0. Dat t = ,t>0. PT <=> 5t - 7t + 2 = 0 o t = 1 hoac t = | 2 (thoa man) Suy nghiem x = 0 hoac x = 1. b) Dieu kien x * 0, dat y = — va chia hai ve cho 4 , ta co: x y PT« 1 + 75 . 1 + 75 — - — es> y = log — - — - 1 = 0<=> 3 o X 1 , 1 + 75 1 , — = l o g — — c=>- = log 2 2 3 1 + 75 « x = log^_ 2 i 3 2X c) Ta co 7 2 - 7 3 . 7 2 + 73 = 1, dat t = (72 + 73") , t > PT<^t+-=4«t -4t+l=0 t o t = 2 + 73 hoac t = 2- 73»x = 2 hoac x = -2. 2 d) Dat t = 2 - , t > 0 thi PT <=> t - -1 = 6 x+vx2 2 2 2 <=> 2t - 5t - 12 = 0. Chon nghiem t = 4. nen x + 7 x - 2 = 2 <=>7 x - 2 =2x 2 2 <=> 2 - x > 0 va x - 2 = 4 - 4x + x <=> x < 2 va x 2 2 2 -BDHSG DSGT12/2- <=> x = 2 1 V i du 5: Giai cac phuong trinh: a) x + ( x - 2 ) = 0 b) V2 V4 (0,125) = 4^2 c) Ul6-x d) / x T l - / x ^ T = v x - 1 3 6 X + $fx~+~l=3 X 3 X 3 / 2 Giai a) B K : x > 0 , x - 2 > 0 < = > x > 2 . V o i x > 2 thi V T > 0 nen PT vo nghiem. b) D K : x # 0, PT o 1 6 7 2 (2~ ) = 2 o 2x 2 3 x 3 7 £ 2x— 6 2 2 =2 k > 2 x 3 1 7 3 » - +- - — = 2 3 2x 3 , 1 » 5x - 14x - 3 = 0 <=> x = — hoac x = 3. 5 c) D K : - 1 < x < 16. D a t u = # L 6 - x , v = 7 x + l thi u, v > 0. lu + v = 3 Ta co he: Dat S = u + v, P = uv u + v =17 » _ 22.23"^ 2 = 2 4 4 thi u + v = ( u + v ) = ((u + v ) - 2uv) - 2u v 4 4 2 2 2 2 17 = ( 9 - 2 P ) - 2 P 2 2 2 2 2 = 2 P - 3 6 P + 81. 2 Do do P = 2 hoac P = 16. V i S - 4P > 0 nen chon P = 2 suy ra S = 3 nen nghiem x = 1 hoac x = 15. 2 d) D K : x < - 1 hoac x > 1. V i 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. P T < » 7(x + l ) - 7 ( x - l ) 2 2 = v x ^lci>6 / Y x+1 x-l x - l Vx+1 e p ^ i . t > 0 thi PT t - - = 1 < = > t - t - l = 0 x-l t 1 + V5 Chon nghiem t suy ra nghiem ciia PT cho la: 2 x=± 't + r 6 VOI t = t-1 1 + V5 V i du 6: Giai cac phuong trinh sau: b) 3 .8 a) 3 * = 4 " 4 X 3 c) S"" ^ =8.4 1 2 X+1 =36 X 5 15 -BDHSG DSGT12/2- Giai a) Hai ve deu duong, logarit hoa theo co so 10 ta co : log 4 « x = log4(log 4) log 3 3 3x x-2 2 = 2 o^ 3TX-2 ~ . 2* = 1 _ 3 o2 . n,2 ,x-2 1 <=> x - 2 = 0 hoac 3.2 =1 4 log3 = 3 log4 <=> x x b) PT o 3 X X+1 3.2 2 3 l X 2 +1 X+1 s » x = 2 hoac 2 = - <=> x = 2 hoac x = - 1 - log32. 3 c) Hai vd deu duong, logarit hoa hai ve theo co so 2 ta co: X+1 log (3 - .2 ) = log (8.4 - ) x 1 x2 x 2 2 2 <=> (x - l)log 3 + x = log 8 + (x - 2)log 4 <=>x -(2-log 3)x + 1 -log 3 = 0 o x = 1 hoacx= 1 -log 3 d) Hai ve deu duong, logarit hoa hai ve theo co so 5 ta co: 2 2 2 2 2 2 2 2 < IM / ^^ i r ^(x-l)l0g (-) -l0g (-)-— 3 1 5 3 + 3 x 4 1 - - 5 O x(log 3 - 1) - log 3 + 1 - I + I log 3 = - 1 - \ 5 5 <=> X 5 4 log 3 - 7 ^ - 4 + l o g ^ ^ 5 _ 2(log 3-4) x 5 2 4log 3-7 5 V i du 7: Giai cac phuong trinh: a) 2 log3x2 .5 c) x l o g 4 +4 log3X l o g x =400 b) 4 =32 d) 3 l n x + 1 -6 l n x -2.3 l n x 2 + 2 I ^ + 3 ' i x—^ V ^ Giai , U B 4 U 6 4 = a) Dieu kien x > 0, phuong trinh <=> 4 log3 x .5 400 = 20 <=> log x = 2 x = 9 (thoa man) <=> 2 0 b) D K : x > 0, PT « 4 . 2 - 6 - 18.3 =0 lnx thi duoc PT <=> 4t - t - 18 = 0. Chia ca hai ve cho 3 , dat t log3X 2 3 2lnx lnx 21nx 21nx 9 _ Chon nghiem t = — <=> x = e • • 2 c) D K : x > 0, ta co: x = 4 ^'° 2 & l o g 4 nen PT « 2.4 logx -BDHSG DSGT1Z/2- Iog = 32 o 4 l o g x e4 =4 l o g 4 l o ^ = 4 x l o g x = 16 o logx = 2 o x = 100. d) D K : x > 0, dat t = log x thi x = 4 l 3 PT o V3 . 3' + -L . 3 = 2 o 4.3' = V3 . 2' V3 1 1 log; ._ _ , V3 o t = l o g ^ Vayx= 4 3 V 73 l-j = ^ Vi du 8: Giai cac phuong trinh: a) (4 - Vl5) + (4 + Vl5) =8 b) 81 tanx tanx sin2 x + 81 c) (cos72°) + (cos36°) = 3.2~ x 30 c x x d) e Giai sin(x—) 4 = tan x a) Vi (4 - y/l5 )(4 + 7l5) = 1 nen dat (4 - JTE ) = t, t > 0 thi phuong tanx trinh ot+ ^=8«>t -8t+ 1 = 0«t = 4±Vl5. 2 Do do tanx = - 1 hoac tanx = 1 nen nghiem x = ±— + leir, k e Z. 4 b) Datt= 81 , 1 t = 27 hoac t = 3 (chon) 2 Do do 3 4s,n2 x = 27 hoac S ^ =3 <=> 4sin x = 3 hoac 4sin x = 1. 4 2 2 ^V3, „ . 1 o sinx = ± — h o a c sinx = ±— 2 • 2 it <=> x = ±— + kn hoac x kn, k e Z. 6 • 3 c) Phuong trinh: (2cos72°) + (2cos36°) = 3 2sin36°.cos36°.cos72 Vi:2cos72° 2cos36° =1 sin 36° x x 0 Dat t = (2cos72°) , t > 0 thi PT o t + ± = 3 x ^ t 2 _ 3 t + 1 = 0 « t = N/5 + 1 Ta co: 2cos72° = 2sinl8° = — — - suy ra nghiem x = ±2. d) Dieu kien cosx * 0, v i sinx = 0 khong thoa man phuong trinh nen PT ,_ %/2sinx 72 cosx 72(stnx-cosx) s m x e * e <=> = cosx sinx cosx A 10 -BDHSG DSGT12/2- Dat u = sinx, v = cosx, u, v e ( - 1 ; 1), u.v > 0 nen ta co phuong trinh 72u Vgv o 2 ~2~ 0 U 72t V Xet ham s6 y = f(t) = —, vai t e (-1; 0) u (0; 1). V2t ^ 72t e 2 V2t v y' = = ———— < 0 suy ra ham so nghich bien tren t 2t moi khoang (-1; 0) va (0; 1). Vi u, v cung dau nen u, v cung thuoc mot khoang (-1; 0) hoac (0; 1) do do PT 2 2 2 <=> f(u) = ftVr'o u = v «tanx = 1 <=>x= — + kn (chon). 4 Vi du 9: Giai cac phuong trinh: a)(sin-) + (cos-) = 1 b) 4 - 3 = 7 5 5 x x X X c)(-) = x + 4 d)2 = x+l. 3 Giai 71 71 a) V i 0 < sin— < 1 va 0 < cos— < 1 do do: 5 5 Neu x > 2 thi ta co (sin- ) < (sin- ) va (cos- ) < (sin- f 5 5 5 => VT < 1 (loai). x x x 2 x 5 Neu x < 2 thi ta co (sin- ) > (sin- ) va (cos - ) > (sin- ) 5 5 5 5 => VT> 1 (loai). Neu x = 2 thi PT nghiem dung, do la nghiem duy nhat 13 b) PT <=> ( — ) + ( — ) = 1 va ta co x = 2 thoa man PT. V i ve trai la ham so 4 4 nghich bien tren R nen co nghiem duy nhat x = 2. c) Vi 0 < - < 1 nen khi x > -1 thi VT < 3, VP > 3 (loai), khi x < -1 thi VT > 3. 3 VP < 3 (loai), con khi x = -1 thi PT nghiem diing. Vay nghiem duy nhat lax = -l. d) PT <=> 2 - x - 1 = 0 Xet f(x) = 2 - x - 1, D = R. Ta co: f '(x) = 2 ln2 - 1, f "(x)= 2 .ln 2 > 0, Vx Do do f '(x) dong bien tren R, f '(x) = 0 <=> x = -log (ln2) x x 2 x 2 x X X X x 2 2 -BDHSG DSGT12/2- 11 BBT X -loKln2) 0 •+ f + 0 +00 f ° + o c Vay f(x) = 0 co toi da 2 nghiem ma f(0) = f ( l ) = 0 nen tap nghiem la: S = {0; 1}. Minh hoa bang do thi cau c) va cau d) y t y V i du 10: Giai cac phuong trinh: a) (^6 + V l 5 ) + ( ^ 7 - v / l 5 ) x c) 6 + 15 = 3 X X+1 + 5.2 x b) (2 - V3) + (2 + V 3 ) = 4 = 13 x x X d) x.2 = x(3 - x ) + 2 ( 2 - 1). X x x Giai a) Ta co x = 3 la nghiem ciia phuong trinh, v i ham so f(x) = ( \ o + V l 5 ) + (^7 - V l 5 ) x x la tong cua hai ham so mu voi co so Ion hon 1 nen f(x) dong bien tren R. Vay x = 3 la nghiem duy nhat ciia phuong trinh. b) Ta co x = 1 la nghiem ciia phuong trinh. Bien doi PT 2-V3V (z + S = 1 thi ve trai la ham f(x) nghich bien, vay x = 1 la nghiem duy nhat ciia phuong trinh. c) P T « 6 - 3 . 3 + 15 - 5.2 = 0 o (2 - 3)(3 - 5) = 0 ci> 2 = 3 hoac 3 = 5. <=> x = log23 hoac x = log35. d) PT » x.2 - x(3 - x) - 2.2 = 0 o 2 (x - 2) x - 3x + 2 = 0 2 (x - 2) + (x - l ) ( x - 2) = 0 o (x - 2)(2 - x - 1) = 0 <=> x - 2 = 0 hoac 2 + x = l<=>x = 2 hoac x = 1. (VI f(x) = 2 + x dong bien tren R va f(0) = 1). X X X X X X x X X x x x X 12 -BDHSG DSGTi?/?. V i du 11: Chiing minh rang phuong trinh: a) 4 (4x + 1) = 1 co diing ba nghiem phan biet x b) x 2 x + 1 = (x + l ) co mot nghiem duong duy nhat. x Giai a) PT « • 4 (4x + 1) - 1 = 0. Xet ham s6 f(x) = 4 (4x + 1) - 1, D = R. x 2 x 2 Ta co f'(x) = 4 ln4.(4x + 1) + 8x .4 = 4 [ln4.(4x + 1) + 8x]. x 2 X x 2 f (x) = 0 o ln4.(4x + 1) + 8x = 0 O (41n4)x + 8x + ln4 = 0 (*). 2 2 PT (*) nay co biet thiic A > 0 nen co diing 2 nghiem phan biet. Tir bang bien thien ciia f(x) suy ra phuong trinh f(x)= 0 co khong qua ba nghiem phan biet. Mat khac: f ( - ) = 0, f(0) = 0; f ( - 3 ) . f ( - 2 ) < 0 Do do phuong trinh f(x) = 0 co diing ba nghiem phan biet: xi = 0, x = — - , x e (-3; - 2 ) . 2 3 b) Voi x > 0, PT <=> (x + l)lnx = xln(x + 1) » (x+ l)lnx - xln(x + 1) = 0 Xet ham so f(x) = (x + l)lnx - xln(x + 1), x > 0. f ( x ) = l n x + ^ i - l n ( x + l ) - — = hi-?+- +x x+1 x + l x x + 1 f "(x) = [ —~) ^-j < 0, Vx > 0 nen f' nghich bien tren (0; +°o), ^ X + x" x ) (x + 1) vi l i m f '(x) = 0 nen f '(x) < 0, Vx Do do f(x) nghich bien tren R nen f(x) = 0 co toi da 1 nghiem. Ma ham f(x) lien tuc tren khoang (0; +oo), f(2) = 31n2 - 21n3 = ln8 - ln9 > 0 va f(3) = 41n3 - 31n4 = ln81 - ln64 > 0 => dpcm. Cach khac: Xet ham f(t) = — , t > 0. Vi du 12: Giai cac phuong trinh: a)2 x + l c) 3 ^ -4 x =x-1 - 2 ^ b) 4 = 4c~^ b g 3 X +2 I o g 3 A =2x d) cot2 = tan2 + 2tan2 x x x+l Giai a) P T « 2 X + I + (x+ l) = 2 2x + 2x. Xet ham sd f(t) = 2 + 1.1 e R thi f '(t) = 2'.ln2 + 1. l V i f ' (t) > 0, Vt nen f dong bien tren R. PT f(x + 1) = f(2x) o x + 1 = 2 x o x = 1. -BDHSG DSGT1Z/2- 13 b) D K : x > 0, dat t = log x => x = 3't t (4) t '2\ =2 + <3) 3 4V (2^ V i f(t) = | J J + | ± J ta co f '(t) > 0 va f(0) = f ( l ) = 0 nen chi co 2 nghiem t = 0 hoac t = 1 <=> x = 1 hoac x = 3. c) Dat t = Vcosx , 0 < t < 1 thi PT o 3 - 2' = t <=> [ -1 + \ = 1. l , y 2 .ln- + l-t.ln3 Xet f(t) = ( | J + - L . _ i < < 1 thi f ' ( t ) = S__ t 9 t Xet g(t) = 2'.ln- + 1 - tln3 voi 0 < t < 1 thi g'(t) = 2 .ln2.1n- - ln3 < 0 3 3 nen f '(t) nghich bien tren [0; 1]. Lap BBT thi f(t) = 1 co toi da 2 nghiem ma f(0) = f(l) = 1 nen PT tuong duong t = 0 hoac t = 1. t <=> cosx = 0 hoac cosx = 1 o x = k27i hoac x = — + kn. 2 d) DK: 2 * k- . Dat t = tan2 thi PT o - = t + o t - 6t + 1 = 0 4 t 1 -1 ot = 3 +2^2 =(V2 ± l) ot = ±(^ + 1) X x 4 2 2 2 2 Vay tan2 = ±{42 ± 1), tu do suy ra nghiem x. Vi du 13: Giai cac phuong trinh sau: a) log [x(x - 1)] = 1 b) log (9 - 2 ) = 10 x X 2 log(3 x) 2 c)——\ + —— = 3 d) 5 /log (-x)=log >^ 5 - 4 log x 1 + log x Giai a) PT <=> x(x -l) = 2x -x-2 = 0<=>x = -l hoac x = 2. b) Dieu kien x < 4. PT: 9 - 2 = 2 o 2 - 9.2 + 8 = 0 o2 =l hoac 2 = 8. Chon nghiem x = 0. 14 5 c) V o i x > 0, dat t = logx thi PT <=> + = 3, 5-4t 1+t <=> 2t - 3t + 1 = 0 <=> t = 1 hoac t = - (chon). > 2 2 2 X x 3-3 2x X X 2 Suy ra nghiem x = 10 hoac x = vlO . d) DK: x < 0, PT o 5Vlog (-x) = log (-x) 2 2 o >g (-x).(5-Vlog (-x)) = 0 2 2 <=> ^/log (-x) = 0 hoac /log (-x) =5ox = -l hoacx = -2 2 14 -BDHSG DSGT12/2- N 2 25 t * - , t * - l 4 V j du 14: Giai cac phuong trinh: a) log x + log (x - 1) = 1 c) l o g ( 3 - l ) . l o g ( 3 - 3 ) = 12 2 b) log x + log x + log4X = 1 d) logx-,4 = 1 + l o g ( x - 1). 2 2 x x+, 3 3 3 2 Giai a) D K : x > 1, PT o log x(x - 1) = 1 o x(x - 1) = 2 <=> x - x - 2 = 0. Chon nghiem x = 2. b) DK: x > 0, PT o (1 + log 2 + log 2).log x = 1 2 3 4 <=> (3 + log 2)log x = 1 o log x = 3 2 2 2 3 +log, 2 Vay nghiem = 2 * c) DK: x > 0 , P T o log (3 - 1)[1 + log (3 - 1)] = 12 Dat t = log (3 - t) thi PT <=> t ( l + t) = 12 o t + t - 1 2 = 0 o t = - 4 hoac t = 3. o log (3 - 1) = - 4 hoac log (3 - 1) = 3 1 hoac 3 - 1 = 27 81 82 o 3 = - hoac 3 = 28 <=> x = log 82 - 4 hoac x = log 28 81 d) D K : x > 1, x ^ 2, PT <=> 1 + l o g ( x - 1) log (x-l) 3+21og 2 x x x 3 3 x 3 2 x x 3 3 X x X 3 3 2 2 Dat t = log (x - 1) thi PT: - = 1 + t « t + t - 2 = 0 2 2 5 o t = l hoac t = - 2 . Giai ra nghiem x = — hoac x = 3. 3 ' V i du 15: Giai cac phuong trinh: a) log [(x + 2)(x + 3)] + i l o g 2 L Z ^ = 2 2 x+3 b) l o g ( x + 12).log 2= 1 4 2 4 x c) - ^ - l o g ( x - 2 ) - ^ = log V 3 x - 5 b d log x l o g 9x d) log 3x l o g 27x 2 1 3 27 9 81 Giai (x + 2)(x + 3 ) > 0 a) D K : x-2 >0 x +3 PT <=> log 4 <=> (x + 2)(x + 3) -BDHSG DSGT12/2: x<-3 x>2 x-2 x+3 = log 16 <=> x - 4 = 16. 2 4 x = 20 o x = ±2 V5 (chpn). 2 b ) B K : x > 0 , x * l . P T o - l o g ( x + 12) — ! — = 1. 2 log x <=>log (x + 12) = log x »x+12 = x ox -x-12 = 0. Chon nghiem x = 4. c) DK: x > 2, phuong trinh tro thanh: 2 2 2 2 2 2 2 ^log (x - 2) + ilog (3x -5)=U log (x - 2)(3x - 5) = 2 D b d 2 cs> (x - 2 ) ( 3 x - 5 ) = 4 o x = 3 hoac x = - Chon nghiem x = 3. 3 d) DK: x > 0, x * i x * —, dat t = log x thi PT o—= 3 27 ' 1 + t 3(3 +1) ot + 3t-4 = 0ot=l hoac t = -4. 1 Suy ra nghiem x = 3 hoac x = 81 Vf du 16: Giai cac phuong trinh: 2 2 2 2(2 + t) 3 & 2 a) l g (4x) + log ^- = 8 2 c) log xlog x + log log x = 2 b) log ^ x+31og x + log x = 2 2 0 2 2 2 1 8 4 2 2 d) l o g 1 6 + l o g 64 = 3 4 x2 2x Giai a) DK: x > 0, ta co log — = log, x - log 8 = 2 log x-3 2 2 2 2 \2 log (4x) = logj 4 + logi x 2 V22J 2 = ( - 2 - l o g x ) = ( 2 + log x ) 2 2 2 2 Dat t = log x thi PT <=> (2 + t ) + 2t - 3 = 8 ot + 6t-7 = 0<=>t = l hoac t = -7. Suy ra nghiem x = 2~ hoac x = 2. 2 2 2 7 b) DK: x > 0, dat t = -log ^ x thi PT » t + -t - it = 2 c> t + t - 2 = 0 2 2 2 2 V 2 <=> t = 1 hoac t = -2. Giai ra nghiem x = —, x = 72 ' 2 c) DK: x > 1, phuong trinh tro thanh log log x + log log x = 2- o ilog log x + log f^log x j = 2 22 2 11 3 o 2 22 2 2 2 2 - l o g l o g x + l o g - + log log x = 2 <=> - l o g l o g x = 3. 2 2 2 2 2 2 2 o log log x = 2 <=> log x = 4 <=> x = 16 (chon). 2 16 2 2 -BDHSG DSGT12/2- d) D K : x > 0 , x * l , x * i thi 2 PT 02102,2 + = 3o + —- = 3 l + log x log x 2 l + log x 2 2 Dat t = log x thi PT: - + —- = 3 o 3t - 5t - 2 = 0 t 1+t 2 2 o t = 2 hoac t = -— Suy ra nghiem x = 4, x = JJ= 3 %/2 V i du 17: Giai cac phuong trinh a) log x . log x = log x + log x b) 21og x . log x + log x - 101og x = 5 c) log x log x log x = log x log x + log x log x + log x log x Giai a) DK: x > 0, ta co x = 1 la mot nghiem. 5 3 2 5 3 5 2 3 2 5 5 2 3 2 5 3 5 Neu x * 1 thi PT o =—-— + —-— log 51og 3 l o g 5 log 3 o log 5 + log 3 = 1 o log 15 = 1 o x = 15 (chon) b) DK: x > 0, PT o log x (21og x + 1) - 5(21og x + 1) = 0 o (log x - 5)(21og x + 1) = 0 o log x = 5 hoac 21og x = - 1 x x x x x x x 2 2 5 5 5 2 5 o x = 32 hoac x = —T= (chon). V5 c) DK: x > 0, phuong trinh o (lgx) = (lgx) (lg2 + lg3 + lg5) o (lgx) (lgx - lg30) = 0 o lgx = 0 hoac lgx = lg30 O x = 1 hoac x = 30 (chon). V i du 18: Giai cac phuong trinh a) log x = 3 - x b) log x + log (2x - 2) = 2 3 2 2 2 3 c) l o g ( l + V x ) = log x 2 4 d) l o g ( l + Vx + Vx) = - l o g Vx 3 3 2 Giai a) D K : x > 0, v i ham so ve trai dong bien, ham so ve phai nghich bien va x = 2 la nghiem nen do la nghiem duy nhat. b) D K : x > 1. Ta co f ( x ) = log x + log (2x - 2) la ham dong bien nen f ( x ) > f(3) = 2 voi x > 3 va f(x) < f(3) = 2 voi 1 < x < 3. Vay x = 3 la nghiem duy nhat. c) D K : x > 0, dat log x = y thi x = 3 3 4 y 3 r— r— f l Y ( P T o l o g ( l + V 3 ) = y « l + V3 = 2 o | ± | + y y R V y 2 -BDHSG DSGT12/2- 17 Ta co y = 2 thoa man phuong trinh, v i ve trai la ham nghich bien nen PT co nghiem duy nhat y = 2 nen x = 2. d) D K : x > 0 , d a t x = 2 t h i P T r 2 y o log (l + 2 + 2 ) = -log 2 <=> log (l + 2 + 2 ) = 4y 3 ' 1 f~ '16^ T ol+2 + 2 = 3 <=> = 1. ,8lJ ,81.J V 8 l j 6y 4y 6y 3 6y 2 6 y 4y 3 4 y r 4y 6 4 + + lY T64Y fiej — +\ — + — nghich 81J \ 81 J I 81 bien tren R nen y = 1 la nghiem duy nhat, do do- PT cho co nghiem x = 2 V i du 19: Giai cac phuong trinh: a) 21og x = x b) l o g [ 3 1 o g ( 3 x - l ) - l ] = x c) log x + log (x + 1) = log (x + 2) + log (x + 3) Ta co y = 1 thoa man va vi ham so f(y) = 1 1 2 2 2 2 d) log 3 3 4 ^ 2x + 4x + 5 x 2 + x + 3 x 2 + 3 5 x + 2 2 Giai a) DK: x > 0, PT: log x =-t> — = — . 2 x 2 2 6 Xet ham s6 f(x) = —. x > 0 thi f(x) = ° X f'(x) = 0 o x = e, lap BBT thi 1-1 x X - In 9 r(2) f(x) == f(4)=i 0 co toi da 2 nghiem ma f(2) = f(4) = — nen S = {2; 4 } . 1/2 + 1 b) D K : 3x - 1 > 0, 31og (3x - 1) > 1 o x > 2 3 x = log (3y-l) 2 Dat y = log (3x - 1) thi co he: 2 y = log (3x-l) 2 Do do log (3x - 1) + x = log (3y - 1) + y 2 2 Xet f(t) = log (3t - 1) + t, t > - thi f' (t) = + 1 > 0 voi moi t > 3 (3t-l)ln2 nen f la ham dong bien, do do phuong trinh f(x) = f(y) <=> x = y o x = log (3x - 1) cs> 3x - 1 = 2 2 3 X 2 o2 -3x+l=0. Xet g(x) = 2 - 3x - 1, x > x X 3 Ta co g'(x) = 2 .ln2 - 3, g"(x) = 2 .ln 2 > 0 nen g* dong bien tren D. Do do g(x) — 0 co toi da 2 nghiem, ma g ( l ) = g(3) = 0 nen suy ra tan nghiem S = {1;3}. x 18 -BDHSG DSGTW2- x 2
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