Cuốn sách Phương pháp giải các chủ đề căn bản Giải tích 12 của tác giả lê hoành phò được biên soạn theo chuẩn kiến thức và kỹ năng chương trình toán giải tích lớp 12. Sách bao gồm các phần tổng hợp các kiến thức cùng chủ đề căn bản về giải tích nhằm giúp các em học sinh nắm vững kiến thức trọng tâm.
515.076 PH561P NGirr.ThS LE H O A N H a ren luyen kl nang lam bai H a No NHA XUAT BAN DAI HOC QUOC GIA HA NOI GIAI PHO NGIfT.ThS LE HOANH PHO GIAI CAC C H U C A N D E B A N (Boi dm!smmmi!mifi cs^G NHA XUAT BAN flAI HOC QUOC GIA HA NOI Ha Nol NHA XUAT BAN DAI HOC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba Trang - Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896: Hanh chinh: (04) 39714899: Tonq bien tap: (04) 39715011 Fax: (04)39714899 * Chiu trdch nhiem xuat * * ban: Gidm doc - Tong bien tap: Bien tap: TS.PHAM T H I TRAM L A N HUONG bdi: N G U Y E N KHCJl M I N H Che ban: N H A SACH HONG A N Saa Trinh bay bia: VO THI T H I T A Doi tdc lien ket xuat ban: N h a sach H O N G A N S A C H LIEN KET CAC CHU DE CAN BAN GIAI TfCH 12 Ma so: 1L- 154DH2014 In 2.000 cuon, khd 17 x 24cm tai Cong ty Co phan VSn hoa Van Lang Giay phep xuat ban so: 463-2014/CXB/10-99 OHQGHN, 14/03/2014 Quyet dinh xuat ban so: 154LK-TN/Q0-NXB OHQGHN, v ^ i > In xong va nop iuu chieu quy II nam 2014 Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 , Idp 12 nam vOng kien thifc can ban ve mon Toan tCr luc vao THPT cho den chuan bi thi Tot nghiep, tuyen sinh Cao dang, Dai hoc, tac gia da bien scan bo sach P H l / O N G P H A P G I A I gom cuon: - C A C C H U D E C A N B A N DAI S O 10 - C A C C H U D E C A N B A N HINH H Q C 10 - C A C C H U D E C A N B A N DAI S O - GIAI T I C H 11 - C A C C H U D E C A N B A N H I N H H Q C 11 - C A C C H U D E C A N B A N GIAI T I C H 12 - C A C C H U D E C A N B A N HINH H O C 12 TL/ nen Toan can ban nay, cac ban c6 the nang cao dan dan, bo sung va md rpng kien thufc va phUdng phap giai Toan, ren luyen ky nang lam bai va tC/ng bade giai dung, giai gpn cac bai tap, cac bai toan kiem tra, thi CLT Cuon C A C C H U D E C A N B A N G I A I T I C H npi dung la phan dang Toan, tom tat kien thac cac chu y; phan tiep theo la cac bai toan chpn vdi nhieu dang loai va mac dp; phan cuoi la hay dap so c6 16 chu de vdi va phUdng phap giai, Ipc can ban minh hpa bai tap c6 hadng dan DO da CO gang kiem tra qua trlnh bien soan song khong tranh khoi nhQng sai sot ma tac gia chaa thay het, mong don nhan cac gop y cua quy ban dpc, hpc sinh de Ian in sau hoan thien hdn Tac gia LE HOANH PHO O CHUD E I T i N H t>ON t>IEU DANG TOAN • TiM KHOANG DONG BIEN VA NGHjCH BIEN Dinh nghia: Hdm so f xdc dinh tren K Id mot khodng, doan hodc nica khodng - f dong hien tren K neu vai moi xi, X2 e K: x/ < X2 =^f(x\) < f(x2) -f nghich hien tren K neu vai moi x/, X2 e K: xi < X2 =>f(xi) > f(x2) Dieu kien can de ham so dffn difu Gid su hdm so c6 dao hdm tren khodng (a; b) do: - Neu hdm so f dong bien tren (a: b) thif '(x) > vai moi x e (a; b) - Neu hdm so f nghich bien tren (a; b) thif'(x) ^0 vai moi x e (a; b) Dieu kien dii de hdm so dffn dieu Gid su hdm so f c6 dqo hdm tren khodng (a; b) do: Neuf'(x) > vai moi x e (a; b) thi hdm so f dong bien tren (a; b) Neu f'(x) < vai moi x e (a; b) thi hdm so f nghich bien tren (a; b) Khi f '(x) = chi tgi mot so hiru hgn diem cua (a; b) thi ket qud tren vdn dung Neu hdm so f dong bien tren (a; b) vd lien tuc tren nita khodng [a;b); (a;bj; doan [a:b] thi dong bien tren nica khodng [a;b); (a;bj; doan [a;b] tuang icng Tuang tu cho nghich bien Phumtg phdp xet tinh dffn dieu: - Tim tap xdc dinh - Tinh dao hdm, xet ddu dqo hdm, lap bang bien thien - Ket ludn Chiiy: 1) Cong thirc vd quy tdc dao hdm y^C =>y' ^0:y=x ^y' = l;y = x" =^y' = nx"-'; ^y=J—(x>0); y= '4^=>y'=— 2Vx n"-47 y = sinx =:>y' = cosx; y = cosx =>y' = -sinx; y= ^ y = tanx =^y = 1r — ; y = cotx =>y , = — -r1— cos X sin X (u + v)' = u' + vV (u-v)' = u'- v'; (u v)' = u'.v + u.v'; u V - u.v ;f'.-f'u.u'x 2) Phuang trlnh luang gidc ca ban: X = a + k27i sinx = sina X = 71 - a + k27i (keZ) X = a + k27i cosx = cos a X = - a + k27r (keZ) tanx ^ tan a c^x a + kTt (k e Z) cotx = cot a x = a + kn(k e Z) Bai toan 1: Tim khoang dong bien, nghich bien cua ham so: a)y = x^-8x + b) y = x^ - 2x^ + x + Gidi a) Tap xac dinh D ^= R Ta c6 y' = 2x - Cho y' = « 2x - = x = 4 +00 Bang biSn thien (BBT) x -00 + y' y Vay ham so nghich bien tren (-oo; 4), dong bien tren (4; +oo) b) D = R Ta CO y' = 3x^ - 4x + Cho y' = « 3x^ - 4x + = o x = - hoac x = +00 1/3 BBT x -00 + 0 + y' ^ ^ y ^ ^ ^ Vay ham so dong bien tren moi khoang (-co; ^) va (1; +co), nghich bi6n tren khoang ( ^ ; 1) Bai toan 2: Tim khoang dong bien, nghich bien cua ham so: a)y-x''-2xl b) y = x'"* + 9x^ - Gidi a) Tap xac dinh D = R y' = 4x^ - 4x = 4x(x^ - 1), y' = x = hoac x = ±1 BBT X -co - y' y -1 + +00 - + +00 +00 ^ - ^ Vay ham so dong bien tren moi khoang (-1;0) va ( ; +oo), nghich bien tren khoang (-oo;-1) va (0; 1) b) D - R Ta CO y' = 4x^ + 18x = 2x(2x^ + 9), y' = x = y' > tren khoang (0; +co) ^ y dong bien tren khoang (0; +QO) y' < tren khoang (-oo; 0) => y nghich bien tren khoang (-co; 0) Bai toan 3: Tim khoang dong bien, nghich bien cua ham so: a)y = 3x-9 b)y = x + - x Giai a)D = R \ Ta CO y' = -6 (l-.r) < vai moi x ^ nen ham so nghich bien cac khoang (-oo; l ) v a ( l ; + o o ) b) TapxacdinhD = R \ Taco BBT: - ^ = ^ ^ - ^ , y ' - < » x = ±V3 x^ X" X -30 y' + +00 -V3 - - + y Vay ham so dong bien tren khoang (-oo; - V3 ) va (V3 ; +oo), nghich bien tren moi khoang (- V3 ; ) v a ( ; V ) Bai toan 4: Tim cac khoang don dieu cua ham so: K^ x+ \ jc +8 2x X -9 Gidi a) D = R Ta c6: y' = -X- y' = -2x + {x'+%f - X - 2x + = X = -4 hay X =2 BBT: X -00 - y' -4 +00 - y Vay ham so dong bien tren khoang ( -4; 2) va nghich bien tren cac khoang -4), (2; + 0 ) (-co; b ) D = R \6 y' = ~^^^ < 0, V x ^ ± (x'-9)' Do y' < tren cac khoang (-oc; -3), (-3; 3), (3; +oc) nen ham so da cho nghich bien tren cac khoang Bai toan 5: Xet su bien thien ciia ham so tren doan, nua khoang: a) y = ^|9-x^ b)y= V x ' - x + Gidi a) Dieu kien - x" > » -3 < x ^ nen D - [-3; 3] Voi -3 BBT: < X < thi y' - X - X ^ , y' = ^ y 0 y' + ^ ^ X = ^ Vay ham so dong bien tren khoang (-3; 0) va nghich bien tren khoang (0; 3) Do ham so f lien tuc tren doan [0; 2] nen ham so dong bien tren doan [-2; 0] va nghich bien tren doan [0; 2] b) V i A' = - < nen x^ - 2x + > 0, Vx => D = R „ , 2x-2 x-1 Fa CO y = —, = , 2Vx'-2x + Vx 2x +7 y'^0 l,y'x< Va f lien tuc tren R nen ham so nghich bien tren nura khoang (-oo; ] va dong bien tren nua khoang [1; +oo) Bai toan 6: Xet su bien thien cua ham so: b)y = a)y Vl6^ a) DK: 16 - x^ > « X' x+2 Gidi < 16 -4 < X < D = (-4; 4) Ta CO 16 > , Vx e (-4; 4) y' = (16-x-)Vl6-xVay ham so dong bien tren ichoang (-4; 4) b) D = [0; +00) Vai x > 0, y' = BBT: +00 - + y' , y' = Vx^ > ^ - ^ = ^ ~ ^ ±1 x^ > hoac x < -1 hoac x > y' y ' > v a y ' = x = hoac x = 2n V i h a m so l i e n tuc t r e n doan [ ; 2n] nen h a m so d o n g b i e n t r e n doan [ ; 2::] b) y ' === - sinx T r e n k h o a n g ( ; n) y' > s i n x < - — < x < — , • r> 7r , _ 571 7t y' < s m x > - < » < x < — hoac — < x < — 6 V a y h a m so d o n g b i e n t r e n k h o a n g (—; — ) , n g h i c h biSn tren m o i k h o a n g 6 (0;5)va(^;H) 6 10 b) Neu a" + c = b thi logb+ca + logb-ca = 21ogb+ca.logb-ca Giai _ a) a' + b^ = 7ab ^ (a + b)^2 = 9ab a +b = Vab => dpcm b) Theo gia thigt: a^ = (b - c)(b + c) Xet a = 1: diing Xct a ^ thi loga(b-c) + loga(b+c) = : log^a log^a 1 =2 nen logb+ca + logb-ca = 21ogb+ca.logb-ca Bai toan 10: Trong khai trien nhi thuc Tim x''^'^"' + ' ^ / , biStsS hang thutubang 200 X? Giai DK: X> 0, X ^ 10 6-* u la c6: _^ EC, + JC' k So hang thu ung vai k = 3, theo gia thiet bang 200 nen: (^3^2(igx.ii Ig" X + =200«x'''^^^' = < » — ^ — l g x = ] 41gx + x = 10 •lgx = l Ig X - = lgx = - DANG TOAN x = 10 (Chon) HAM SO LUY THlTA, MU, LOGARIT Ham so luy thiva y ^ x": Ham so y = x" dong bien tren (0; +oo) a > 0; nghich bien tren (0; +oo) aT-'» 00 a > ; hm a = < a < Dong bien tren R neu a> I, nghich bien tren R neu 0< a< I Dd thi luon cdt true tung tai diem (0; 1), nam a phia tren true hodnh vd nhdn true hodnh Idm tiem can ngang 202 Ham so logarity = logax: Lien tuc tren tap xac dinh (0; + oo), nhdn moi gid tri thuoc R limlog, x = Ham so y +cokhia>l -00 ; l i m log, X - c o k h i O < a < l ^-^1' a > = + 00 k h i < a ) neu < a < Do thi luon cat true hoanh tgi diem (1: 0), nam a hen phai true tung vd nhdn true tung lam tiem can dung Cdc gi&i lim han: = e'-\ lim , ln(l + x) _ = ; lim jr->0 • r->0 =1 x Do thi va quan he ddixirng: y - logaX y = logaX Bai toan 1: T i m tap xac dinh cua cac ham so sau: a) y = (x^ - 4x + 3)"-b) y = ( x ' - x + 3)^^' 203 Gidi a) Ham so xac dinh k h i : x^ - 4x + ?t x ^ va x VayD = R \ { l ; } b) Ham s6 xac dinh k h i : x ^ - x + > < = > x < l hoac x > Vay D = (-oo; l)u(3;+oo) B a i toan 2: l l m tap xac dinh cua cac ham so sau: b) y = a)y=lg(x'-9) log,(4x-l)-l Gidi a) D K : x^ - > « b) D K : x < -3 hoac x > Vay D = (-oo; -3) u (3; +QC) (ham nghich bien) 4x-ll 4x-l>0 4x-l>0 - < x < ^ V a y D = ( - ; 4 -] B a i toan 3: Chung minh cac giai han: a) lim^^ x->0 b) H m l ^ M ^ _ L ^ = hi a In a X X Gidi a) l i m ^ ^ = lim^ -1 = Hm •f-^o X In a , l o g J l + x) , ln(l + x ) b) hm—^^^^ = limlog^ e In a = In a In a B a i toan 4: T i m cac giai han sau: a) lim 2x e —e x-*0 _„5x b) H m ^ X — x->0 X Gidi a) l i m = -3e".Hm = hni X->0 X b) l i m ->-0 '^^o X = -3e^ 3x = - = -3 = Hm x->0 X—»0 ^ X X , B a i toan 5: T i m cac giai han sau: sin4x a) h m ^-0 3^+5^-2 204 Giai 2^-1 5^-1 X _ In2 + ln5 _ j n l O a) l i m _ = Jim— '•^'3^ +5^ - 3^ - — 5" - " In3 + ln5 " InlS + — X X sin 4x b) l i m ^ ' " ^ ^ = l i m — r x^oe^^-y 7^-1 x Bai toan 6: T i m cac giai han sau: 3-ln7i X , , ln(l + 3x) a) l i m — x-»0 X ••^o - c o s x Giai = J v^O X x->0 l-cos2x 3X 2sin"x -lim 31n(l + x - ) fsinx^' 3x^ Bai toan 7: T i m cac giai han sau: 6"" 4x a) l i m log.O + Sx) -3' b) l i m ^-^0 ln(l + x ) - l n ( l + 3x) Giai y 4x , 5x 4, ^ a) h m = hm = -ln3 l o g , ( l + 5x) 51og,e ^ - o i n ( l + 5x) 6^ 3^ 6^-1 b) l i m =lin'^-*ln(l+6x)-lnO+3x) 3^-1 ln(l+6x) ln(l+3x) - (In6 - ln3): (6 - 3) = ^ l n Bai toan 8: Tim cac giai han sau: a) l i m + -x-3 b) l i m ^ x + 3^^ VX+ y Giai x-3 a) lim 1+x-3; = lim 1+x-3 = e =e 205 X b) lim X ~ lim + — = lim 1+ t->4«0 2J X+ \ Bai toan 9: Ve thj ham so y = f(x) = 2" Suy thi cac ham so y = 2" - 1, y = \ = - \ = ( ^ r , y = 2'''L Gidi y = f(x) = 2\ = R Hm y = +00, X - • K lim y = => TCN: y = (khi x x-> -co) -CC y' = 2" In2 > 0, Vx nen ham so dong bien tren D = R BBT X -00 +00 + y' y —• —' +00 Chox = =>y== = -1 : X = X y=2 Ta c6: y ^ 2^ - = f(x) - 1: Tinh tien thi f(x) xuong duai don vi y = 4.2^ = 2^"^ = f(x + 2): Tinh tien thi f(x) sang trai dorn vi y = -2" = -f(x): Lay doi xung thi f(x) qua Oxy = ( ^ ) " = 2"' = f(-x): Lay d6i xung thi f(x) qua Oy y=2 = f{ X ) ham so chan x > thi y = f(x) nen lay phan va lay doi xung ciia no qua Oy Bai toan 10: Ve thi ham so y = f(x) = logax Suy cac thi ham so y = log22x, y = lg2(x - 3), y = log2(-x), y = log, x , y = log2 Giai y = f(x) = log2X, D = (0; +a)) lim y = +CO, lim y = -oo TCD: x = (khi x 0^) - > 0, Vx > ncn ham so dong bien tren (0; +oo) xln2 y' = 206 BBI X -00 +00 y' -f y Cho X -DC = y = -l x = l =>y = 0, x = = > y = l Ta c6: y = log22x = f(x) + 1: Tinh tien thi f(x) len tren don vi y = log2(x - 3) = f(x - 3): Tinh tien thi f(x) sang phai don vi y = log2(-x) = f(-x): Lay doi xung thi f(x) qua Oy y = log i x = -f(x): Lay doi xung thi f(x) qua Ox y = log2 x =^ f( x ) la ham so chan, x > thi y = f(x) nen lay phan va lay doi ximg cua no qua Oy DANG TOAI\ D A D H A M VA K H A O S A T H A M S O Dao ham (x")' = ax"-', (u")' - au"-'u'; (X nVx"-' > 0), ' u = , vai u> nVu^ (a'y^ cflna; (e')'=- e"; (a")' - a"ulna; (e")' = e"u' Oogax) = - ^ ; xlna ( l n x ) = - ; (ln|x|) = X X Oogauj' = -f• (In u)' = - ; (ln|u|)' = - ulna u u Khdo sat ham so: xel tinh dan dieu cue tri - Gia su ham so f c6 dgo ham tren khodng (a: h), do: Neuf'fx) > vai moix e (a; h) thi ham sof dong bien tren (a; b) Neuf'(x) < vai moi x e (a; b) thi ham so nghich bien tren (a; b) Khif'(x) = chi tgi mot so hihi hgn diem cua (a; b) thi ket qud tren vdn dung - Cho y f(x) lien tuc tren khodng (a:b) chita XQ, CO dgo ham tren cdc khodng (a:xo) vd (xo;b): Neu f '(x) doi ddu tit dm sang ditang thi f dgt cue tieu tgi XQ Neuf '(x) doi ddu tit dmmg sang dm thi f dgt cue dgi tgi XQ 207 - Choy =f(x) CO dao ham cap hai tren khodng (a;b) chica xo: Niu f '(xo) = Ovaf "(xo) > thif dat cue tiiu tai XQ Niu f '(xn) = vaf "(xo) < thif dat cue dgi tai XQ Bai toan 1: Tim dao ham cua cac ham so sau: T -2 b ) y = ^2^+2"" Gidi c) y = - 5'^ + x^ a) y' = e^' + (x - l).2e^' = (2x - l)e^' b) y' = (2^ ln2 + r ^ ln2)(2^ +2-^)-(2'' -2^X2^ ln2-2-^ ln2) (2^+2'^r (2^+2")'-(2^-2'^)-, = {2^+2'^)- ; ln~ = 41n'2 ~ {2^+2-^f c) Ta CO y = x' - 5' + x^ = x" - 5" + e"'"" nen y' = 5x^ - S'lnS + e"'"^(inx + 1) = 5x^ - S^lnS + x^lnx + 1) Bai toan 2: Tim dao ham cua cac ham so sau: a) y = (3x - 2)ln^x b) y = ln(x + V x ' + a ' ) c) y = ^"^^ ^-^ x Gidi a) y' = 31n\ 1+ 2(3x-2)lnx b)y' = Vx' +a' X + A/X" +a^ Vx" +a' ^ , ln(x-+l) c) y = - T — r ^ —• X +1 X" Bai toan 3: Tim dao ham cua cac ham so sau: a)y= ( x + I f - t a n e ' ' b) y = Vln'Sx = c)y = + x' )ll-x-' Gidi a) y' = 2Tt(2x + 1)""' - (1 + tanV)e' (In' Sx) 31n'5x b) y' 5'4\n'-5x S^hr5x 5\/(ln'5x)' ' 6x' c) Dat u = ^ thi y' = — ^ = va u' = 1-x' 3V^ u'Vu 3u nen y 2x- J l + x' 1-xMl-x-^ 208 Bai toan 4: Chung minh: a) Neu y = In thi xy' + = e'' 1+ x b) Ngu y = e^" + 26'" thi: y"' - 13y' - 12y = Gidi \ , - X , a) y = x + suy' xy' + = x + + = x + b) y' = 4e'"- 2e", y" = lee'" + 2e", y"' = 646"" - 26-" nen: y'" - 13y' - 12y = (646'" - 26-") - 13(46^" - 26"") - 12(e'' + 26"") = Bai toan 5: T i m dao ham cap n cua ham so a)y = 5'''' b)y = ln(x-5) c) y = ln(6x^ - x - 1) Giai a) y' = (]cln5).5''\" = (kln5)l5'''' Ta chung minh quy nap: y*"* = (klnS)" S""" b) Vai 5: y = X ^ -1- ; y"= —^1^; x-5 1.2 y"= ((xx s5y (x-5)-^ T u' • u (n) ( - ll ) " " ' ( n - l ) ! la chung minh quy nap: y = — - (x-5)" c) Voi X < hoac x > ^ : y = ln((2x - l)(3x + ) ) = In 12x - | + In 13x+l 1 y ' = x - l + •3x + l \(ni) Ta chung minh quy nap (-irmla" ax + b („,_(-l)"-'(n-l)!2"-' Suyra y'"' = (ax + b) m+l , (-ir'(n-l)!3"-' • + - (2x-l)" (3x + l)" Bai toan 6: T i m khoang den dieu va cue tri ham s6: a)y = b) y = x l e - " Gidi a)D = R \ { } , y ' BBT X y' e^(x-l) , y- = 0 -co - « +00 1 y X = +00 + ^ ^ " ^ -00 209 Vay ham s6 nghich bien cac khoang (-00; 0) va (0; 1), dong bien tren khoang (1;+00), datCT(l;e) = R, y'= (2x - x^)e\' BBT X -00 0 y' y = X = hoac X = + +00 -1-00^ -00 ' Vay ham s6 d6ng biln khoang (0; 2), nghich bien cac khoang (-00; 0) va (2; +00), dat CD(2; 4e\; 0) Bai toan 7: Tim khoang don dieu va cue tri ham so: b)y = x - l n ( l + x) a)y = I n ( x ^ - l ) Gidi a)D-(-cx);-l)u(l;+a,),y=-^ x' - Khi x < -1 thi y' < nen ham so nghich biSn tren (-00; -1) Khi X > thi y' > nen ham so dong bien tren (1; +00) Ham so khong c6 cure tri b ) D = (-l;+cx)), y = l - 1+x , y' = a^' N NiuO a^ M h' thi: logaE > logaF F> NiuO logaF 0 va log4 ^ ^ ^' Jogs^ > log4 ^ b) Ta CO log6l,l > nen 3""^''' > 3° = ( v i > 1) valog60,99 < nen 7""^'°'' < 7° = ( v i > 1) Suyra 3"'^''> 7log60,99 211 Bai toan 4: So sanh cac so: a) 31og2 + log3 vai 21n5 b) log, - va log3 5 Giai a) 31og2 + log3 = log(2^ 3) = log24 < log25 = 2log5 < 21n5 2 b) V i - < va - < nen log, - > log, = 5 3 V i - > va - < nen log3 - < log, = ^ , Tu suy log, — > log, - - ^ ^ Bai toan 5: Chung minh: a) log23 > log34 b) a"' + b"" < c"", n^u m > 1, a + b = c, a > 0, b > Gidi > log34 log32.1og34 < 1: Dung vi log, a) log23 > log34 Vlog, 2, log, < i (log32 + log34) = ^ log3(2.4) < ^ logjQ = b) a Ta CO a"" + b"" < c*" » + b , b > O n e n O < - < l , < - < l c c m a Suy vai m > thi — c Tu ta c6: < — c c b a — , Bai toan 6: Chung minh cac bat dang thiic sau vai moi x > •> a) e > X+ b)ln(l + x ) > x - X" Gidi a) Xct ham s6 f(x) = e' - x - 1, x ^ thi f (x) = e" - > 0, Vx > nen f dong bien tren (0; +oo) vi f lien tuc tren [0; +oo) nen f dong bien tren [0; +oo): x > => f(x) > f(0) = 0: dpcm 212 b) BDT: l n ( l + x ) - x + — > 0, V x > Xet f(x) = l n ( l + x) - X + — , X> 0, f ' ( x ) - ^— > +X va f lien tuc tren [0; +oo) nen f dong bien tren [0; +oo) Do do: X > => f(x) > f(0) = 0: dpcm Bai toan 7: Chung minh: a) 4™^+2'='"^ > V ^ , V x e (0; | ) b) 2=" + vai a > b > Giai a) Ap d u n g bat d a n g Ta can Xet Chung minh: f(x) thuc Cosi: 2^^'" " ^ "^^^ ^ > 2^""' « 2sinx + t a n x > 3x = 2sinx + t a n x - 3x, < x f (x) = 2cosx + — \ - < ^ > 2cos\ — ^ - > V2 - > COS" X COS^ X nen f dong bien tren fO; - ) : x > =^ f(x) > f(0) = 0: dpcm b) Vai a > b > 0, bat dang thuc tuang duang: 4Mn4 + 4' Mil!! a ^ b x - l n ( l + 4') = - ( M n ' - ( l + 4'').ln(l + ' ' ) ) < X" nen f nghich bien: a > b > => f(a) < f(b): dpcm Bai toan 8: T i m gia tri Ion nhat va gia tri nho nh4t cua ham s6 a) f(x) = X - e^' nen doan [-1; 0] b) f(x) = ln(x^ + x - 2) tren doan [3; 6] 213 Giai a) Ta c6: f'(x) = - 2e^^ f'(x) = o x -•Ir In V2 e (-1; 0) l + ln2 f(0) l f(-l) = - l - e - ^ f ( - l n V ) = - Sosanhthi: maxf(x) = f(-lnV2) = - l n V - - , minf(x) = f ( - l ) = - l - e-2 xe[-l,0| b) Ta 2x + l CO f ' ( x ) = - ^ nenf '(x) > 0, Vx e [3; 6] do tren doan [3; 6] + X - X" ham so f(x) dong bien Vay niin f (x) = f (3) = In 10 ; max f (6) = in 40 xe[3,6] Bai toan 9: Cho p > 1, q > thay doi thoa p + q = pq va a, b > a'' b'' Tim GTNN cua T = — + P q ab Giai Xet ham so f(a) = p b" • + - - ab vol a > f'(a) = aP"' - b, f'(a) = 0 a^"' = b « a = b""' Ma p + q = pq o (p-l)(q-l) = nen a - b''"' Lap BBT thi T = f(b''"') = Bai toan 10: Tim cac gia tri cua m de phuonag trinh Vx' +2x + - Vx + = m c6 dung mot nghiem Giai Dat t = Vx + ^ 0, phuong trinh tro V t S ^ - t = m (*) Nhan xet ung vai moi nghiem Ichong am ciia phuong trinh (*) c6 dung mot nghiem cua phuong trinh da cho, do phuong trinh da cho c6 diing mot nghiem ichi va chi phuong trinh (*) c6 diing mot nghiem khong am Xet ham s6 f(t) = V t ^ - t voi t > 0, f'(t) = , ^ - KO Ma f(0) = V3 va lim f (t) = nen c6 bang bien thien: t - f'(t) f(t) l u bang bien thien suy cac gia tri can tim cua m la < m < Vs 214 Bai toan 11: T i m m de pliuomg trinh sau c6 nghiem thuc: 3^f^ + mVx+T = V x ' - l Gidi Dieu kien x > x-1 x-1 Dat i x + =^t = iV x + r.- 1-n e n O < t < V x+1 = x-1 Phuong trinh da cho: I x+1 +m=^ ^ ^ hay la-3t^ + 2t = m x+1 Xet ham s6 f(t) = -3t^ + 2t, < t < Ta c f ' ( t ) = -6t f , f ' ( t ) = o t = ^ t BBT - f 1/3 + f ^ - ^ - Vay phuong trinh da cho c6 nghiem -1 < m < ^ Bai toan 12: T i m m de phuong trinh sau c6 dung nghiem: m V2X+V2X+2V6-X+2V6-X =m Gidi Xet f(x) = V x + V x + V - X + V - X , < X < Vai < < 6, ta c6: X f'(x)f(2) = 0, BBT: X f 1 X > + =^ f'(x) > 0, X < V2x •\/6-x ^ =^ f'(x) < + f T a c f ( ) = 2V6+2V6,f(2) = + V , f ( ) = V l + 2V3 Vay dieu kien c6 nghiem phan biet: 2V6+2V6 < m < V + 215 BAI TAP T O N G HOP Bai tap 1: Dcm gian bieu thuc dieu kien xac dinh: N = a-b V^-Vb a+b Va^ + V b ' a-1 M= r a" I V^ + Va^ Vra=+ a" +1 HD-DS Ket qua N = 2\[ab ; M = -Ja v21 Bai tap 2: Trong khai trien nhi thuc , tim he so ciia so hang vVb^VVb chua a va b CO so mii bSng HD-DS K i t qua C', = ^ = 293 930 9!12! Bai tap 3: Tinh gon a) log5log5 V , n dku can b) log72 - l o g ^ + log Vl08 256 HD-DS a) Ket qua -n b) Ket qua o g - | l o g Bai tap 4: a) Tinh ln6.25 theo c = ln2, d = ln5 b) Tinh Iogi4o63 theo a = log23, b = logaS, c = log72 HDDS a) Ket qua 2d - 2c b) Ket qua lac +1 abc + 2c +1 Bai tap 5: Tim cac giai han sau: a) lim Vl + 2x -VTHOC X-+0 b) lim tanx r ln(l + x ' ) HD-DS a) Ket qua b) Ket qua - ^ 12 Bai tap 6: Chung minh cac bdt dang thuc sau a) e" > + X + — vai moi x > 216 ... GIAI TfCH 12 Ma so: 1L- 15 4DH2 014 In 2.000 cuon, khd 17 x 24cm tai Cong ty Co phan VSn hoa Van Lang Giay phep xuat ban so: 463-2 014 /CXB /10 -99 OHQGHN, 14 /03/2 014 Quyet dinh xuat ban so: 15 4LK-TN/Q0-NXB... HOC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba Trang - Ha Npi Dien thoai: Bien tap-Che ban: (04) 39 714 896: Hanh chinh: (04) 39 714 899: Tonq bien tap: (04) 39 715 011 Fax: (04)39 714 899 * Chiu trdch nhiem... thi f(x) = x''(x' ~ 1) + 3x^(x^ - 1) + > 0: v6 nghiem 2x3 Xet < X < thi f(x) = x'^ + (1 - x^" > 0: v6 nghiem 22 Xet < thi: f'(x) = 13 x'^ - 6x^ + 12 x^ - 6x = 13 x'^ - 6x(x - 1) ^ > nen f ddng bien