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Thực hành luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học: Phần 1

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Tài liệu Luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học gồm 20 chủ đề trọng điểm theo cấu trúc đề thi của Bộ Giáo dục và Đào tạo, 28 bộ đề thi các khối A B, D của trường chuyên Lê Quý Đôn - Đà Nẵng dùng để thi thử qua từng năm để đánh giá học sinh. Mời các bạn cùng tham khảo phần 1 tài liệu.

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Nội dung Text: Thực hành luyện giải đề trước kỳ thi Đại học 3 miền Bắc - Trung - Nam Toán học: Phần 1

  1. Nh^ gi^o uu tij - TTi^c siToAn hpc NGUYEN VAN THONG (Chu bifen) Gl^ng vi«n chfnh - Thac sITo^n hpc NGUYEN VAN MINH Ph6 gi^o su - mn si: NGUYEN VAN HIEU ••a LIIYeNGIIUIliniUIICKVTHIIilUHOCSMliN TOAN HOC ii@® Ddnh cho hpc sinh 12 luy^n thi BH-CB ^ Biin so$n theo n0i dung djnh hudng ^ i mdi cua BO Gi^o Due & B^o Tao THLT MIEN TiWH BINH THUAN MZ DVL J My^6' /I5 I NHAXUJTBiiNTONGHQPTHiiNHPHOHOCHlMINH
  2. I 'K^mdt m MJOl C«0 mill inx-r z^r TWM I \ F M Left noi ddu TRONG DI^M Chiing toi la cac anh em ruot thjl, muo'n vie't quyen sach nay cho the he sau on tap de chuan bi thi v^o dai hoc, vi bU'dc v^o triTdng dai hoc, ngtfcfi hoc sinh b^l dau mot ddi song mdi vk c6 tiTdng lai tiTdi sang khi chpn dtfdc mot triTcJng dai hoc tot, ch^c ch^n r^ng triT^ng nay se chon diem cao. NhiT vay trong qua trinh on GI(3l HAN vA TfNH UfeN TUC C6A M O T H A M S6 luyen cac em can mot tai lieu tUcfng thich. Dieu nay se diTdc thoa man neu cAc em chiu kho on tap theo cac chuyen de mh chung toi bien soan, ch^c ch^n thi se dai I . T 6 M TAT L I THUYET diTpc diem cao. 1. C a c d j n h If ve g i d i h a n ••1 ( ' Quyen sdch nay gom 20 chu de trong diem theo cau true de thi cua Bo giao Gia sur lim f ( x ) = L, va lim g(x) = L j , khi do X-^X() X-+X,) due hang nam, cac vi du duTa ra ttfcfng doi kh6 va c6 hiTdtng dan giai. Tie'p theo la 28 bp de thi cho cac khoi A, B, D cua tri/cfng chuyen Le Quy Don - Da N^ng lim [f(x) + g ( x ) ] = lim f ( x ) + lim g(x) = L | + L 2 x->X() "->"() "-^^o dilng de thi thijr tiTng nSm qua 66 danh gid diTdc hpc sinh, nit ra kinh nghiem lim [ f ( x ) - g ( x ) ] = lim f ( x ) - lim g(x) = L , - L 2 giang day. TriTdng Le Quy Don - Da NSng hang nam deu c6 ti 16 dau vao cac x->X() "-+"() x-^xo trirdng dai hpc la 100%. lim [ f ( x ) . g ( x ) ] = l i m f ( x ) . lim g(x) = L|.L2 Rieng nam 2010 c6 so hpc sinh dat diem cao diTcfc xep thu" 4 trong bang xep x^X() x-^xo ''->X() ='--'v,V' hang cua Bp giao due va Dao tao lim f(x) , f(x) X^X() Cic chuyen de chiing toi deu soan tilf can ban den nang cao, dp kho dii de lim _ ^1 vdi cho cac hpc sinh khd va gioi tU' luyen, neu thong hieu taft ca, ch^c ch^n ring cac X-^X() g(x) lim g(x) L2 X-*X() em se giai du'pc de thi Dai hpc mpt cdch de dang, nh\i chiing toi da tilfng luyen Neu 3s > 0 sao cho f ( x ) < g(x) V x € (xo - e; x,) + s) va ton tai 1 , , rat nhieu em do thu khoa cic tru'dng Dai hpc danh tieng. Dieu quan trong hdn lim f ( x ) , lim g(x) thi lim f(x) < lim g ( x ) nifa vdi each vie't ciia mOt gido vien day chuyen Todn lau nim, nen each gpi x->X() ''-»X() x^XQ x->X{) li • md, d i n d^t mang dam n6t t\i duy nang cao, hpc sinh se diTcJc phat trien kha a. NguyinIfgidinank?p: nSng Toan hpc khi hpc quyen sdch n^y. Neu 3 E > 0: f ( x ) < h(x) < g(x) V x e (x„ - E ; X„ + e) \} Mong muon ciia chiing toi l ^ m the nko de c^c em tif hpc to't mon Toan, va va lim f ( x ) = lim g(x) = L thi lim h(x) = L . kha nang thi dau v^o cdc tri/dng Dai hpc Idn phai nh5 vko sir kien tri cua cic x^X() x-»X() x-»X() em "c6 cong mai s^t c6 ng^y nen kim" cd nhan day that diing vay. b. Cdc dang gidi han dQc bi$t Quyen sich nay cung 1^ ky nipm 3 nim cic em hpc Toan vdi thay: Bich Lien, PhiTdng Thao, Anh ThiT, Mai HiTdng, Thiiy Hien, Th^o Uyen. Cic em da lim = 1 ; lim (1 + x ) x = e X->X() X X^XO I,,,,,.,.,.;,.,: giiip thay chinh sufa ban thao, mpt cich nhipt tinh trdch nhipm. Chiic c^c em se th^nh cong my man trong ky thi s^p idi e^-l ln(l + X) lim = e ; lim = 1 ; lim = 1 Chu bien: Nguyen VSn Thong X->X() x-»X() X x->X() (To trirdng To Todn trirdng chuyen Le Quy Don - Da NJng) 2. G i d i h ^ n d a n g v d d j n h : — 0 Nhd sdch Khang Vi$t xin tran trgng giai thi^ tai Quy dgc gia va xin Idng nghe tnoi y kien dong gop, de cuan sdch ngdy cdng hay han, bo ich hart. P(x):dathu'c, P(x„) = 0 > P(x) Thu xin giH ve: .„,.^-.. _ a. Dang 1. I = lim — ^ vdi
  3. . J I IWtg, „ |,W C6ngiyTNi:ii : ; . W / / A / . U ; . ^ ,UI P,(x) = ( x - X o ) P 2 ( x ) Ne'u Pi(x„) = Qi(x,)) = 0 thi phan tich tie'p • • Ng'u p < q thi ton t a i g i d i han Q,(x) = ( x - X o ) Q 2 ( x ) 1 0>t • N e u p > q t h i khong ton tai g i d i han 0 Qu^ trinh khuT dang v6 dinh ^ la qua trinh k h u r cac nhan tuT chung ( x - X o ) ' ' s e 4. Gi6i han dang v6 dinh « - oo difng l a i k h i nhan du'dc g i d i han xac dinh ttfc la Qk 5^ 0. Phifdng phap: B i e n d d i diTa ve dang g i d i han — ' Khid6I= l i n . - ^ = l i m A W ^ M ^ >'^''()Q(x) x ^ x „ Q , , ( x ) Qk(xo) T i m g i d i han sau ' f ( X ( , ) = g(x„) = 0 O O . (x + V x j - x b. Dang 2. l i m vdi \ lim Vx + V x - V x = l i m . = lim x->xo g(x) f ( x ) , g ( x ) chtfa can thtfc dong bac x-++^x + ^/^+^/x " PhiTdng phap: SuT dung cdc hKng d^ng thtfc de nhan l i e n hdp 6 tuT va mau nh^m true cac nhan tijr (x - x,,) ra k h o i can thtfc. 5. Gidi han dang v6 djnh oo . 0 , . Phifdng phap: Du'a ve dang v6 djnh — 00 ~. ^ ' J A-B mil tioub fi?. i m C h i n g han, t i m g i d i han x 1 A + B lim Vx^ + l - x = l i m - - lim X->+oo . X-»+a> 2 Vx^ +1 +: l + -^+l X A-B 6. Gidi h^n dang vd djnh ham lUging giac PhtTdng phap: Suf dung cac ket qua g i d i han cd ban sau dSy: 2"M/A+2n+^^ A + B ,. sinx , ,. X , • hm = 1; h m =1 x->0 X x->() sinx V sin ax sin ax ,. sinax ,. sinax c. 5.1 = hm vdi • lim = lim I = a. h m a => h m =a ' x^xog(x) [ f ( x ) chtfa can khong dong bac x->0 X x-»0 ax x->() ax x^O X „ r sin ax sinax ax smax a , . ^^"0 g(x) g(x) h m = - ^ = ^^lim .g(Xo) = 0 — iUm x^XQ mil iiii .—:—-— — — —• 11111 = — -—( a , b e R * ) Bi6n doi x->()sinbx x-»()bx sinbx b x-+()sinbx b bx '?yu(v)-ci-rjyv(x)-c •^uW-c 5!/v(x)-c tan ax a sinax ,. tanax 1= lim - - lim lim hm . = a => h m =a x->xo g(x) X-+XO g(x) g(x) x-»o x x^ocosax ax x->0 X D e n day cac g i d i han diTdc tinh theo dang 2 . tanax tanax,. ax ax a ,. tanax a • Hm = hm — • =- hm 3. Gi6i hain v6 djnh — x->otanbx x->{)bx tanbx b x->()tanbx b 00 bx PhiTdng ph^p: X 6 t I = lim ^ v d i P(x), Q(x) la cac da thi?c hoSc cic ham Sinax x-»xo g(x) sinax , sinax ,. ax . a x ^ daj s6. G p i bac P(x) = p; bac Q(x) = q v^ m = min(p, q ) , k h i d6 chia ca va hm hmcosbx. = hm—.cosbx. . . mau cho x " ta c6 k e t luan sau: x-*() tanbx x->() sinbx x->obx sin Dx b bx
  4. Luyfn gidi di truOc kp ihi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyin Van ThOng Hifdng din giai 7. Gidi han dang v6 djnh 1° Phifdng phap: Ta CO lim '^^^^ = ^JZ^ =1 > x^3 (X-3)X x-»3 X f 1 a. Sijfdung: lim(l + x ) " = e ; lim 1+ - = e ,,,, . v l ^ - 1 I- x->() x^+«> Bai 3. Tim hm x->0 X^ b. Xct lim uCx)"*"' CO dang T t'l I-.,, " US;' fili'd i&ili ffH'T X^X() • • Hifdng d§n giai - -- (l + x 2 ) - l • ''1 lim (u-ir X \ •+ X Ta c6: lim — - == lim iim , , = — x-.(.^2 Bie'n doi lim l + (u-l)"-' 3/^1777+^/i77^^1 ''-*"^(i+x2)+^/iT^+i 3 X-»X() 8. Gidi han d?ng v6 djnh cua ham mu va Idgarit: ^ ^^^ ' * Bai 4. Tim hm x-»() X Hi/dng d i n giai Phifdng phap: S^rdung l i m ^ ^ = 1; lim ^ =1 Ta C O x-»() X x->() X II. B A I T A P M I N H H Q A ,. (2N/rT7-2) + ( 2 - ^ / 8 ^ ) ,. X hm lim x-»() X x^O Bai 1. Tim gidi han lim f ( x ) , vdi r(x) = X->1 x^-1 ,„;.,. fijsi Htfdng din sial 2x = Hm Vs-x^ - 2 \/x^ + 7 - 2 ^ x-^o >/l + x + l ) X[4 + 2 W ^ + N/(8-X)^ Ta CO lim (1) x^l x^-l x^-l Matkhac: = hm x->() ^ / ^ ^ + 1^4 + 2 ^ +^(8-x)2J ""12 12 1-x-^ - f x ^ + x + l) hm = lim - = lim \ ' (2) l(X) . (x + 1)'"" + (x + 2)'"*' + ••• + (x + 9)'"" + ( X +100) X ^ l X -1 " - l / v ^ X->1 Bai 5. Tim lim X-»+oo x'"" + 10x'" + 100'" Hi/dng din giai Vx" + 7 - 2 x^-l Va lim = lim ^ ,^100 ^ ^^100 ^ j ^ ^ N 100 x->l X -1 "->! .100 2 ^"'(x^-lWx^+vf+2^/?77 x^+7+4 1+ + ...+ 1 + Ta c6 lim - = 100 X-»+QO 1 .100 19^ loo^ll^ = lim (3) + ^ 1 0 + ^HX. x^l 12 3 v2 x^ +7 + 4 •a x^+7 Bai 6. Tim gidi han lim m , (m, n e N * , m n) , , x-»l 1-x"" 1-x" Thay (2), (3) vao (1). la difdc L = - - - — = - — 8 12 24 Hi^dng dan giai . . ™ ,. x'^ - 4 x ^ +4x - 3 B a i 2 , Tim l i m — lim m f_n. i_^] :; "-^^ x--3x X->1 Al-x'" l-xj
  5. Luy(n gUU d6 trade thi DH 3 miin Bdc. Trung. Nam Todn hoc - Nguyen Van ThOng m - ( l + x + x^+... + x'""') n - ( l + x + x^+... + x""') a a lim 2sm rcos ...cos-^.cos— 2n-l 2 " ' 2^ 2 X-+1 l-x"" 1-x" 2 sin 2" ( l - x ) + ( l - x ^ ) + ... + ( l - x " ' " ' ) ( l - x ) + ( l - x ^ ) + ... + ( l - x " - ' ) lim x-»l (1 - X)(l + X + x^ +... + x " ' " ' ) (1 - x ) ( l + x + x^ +... + x"~') sma 2sin—cos— l + ( l + x) + ... + ( l + x + ... + x " ' ' ^ ) l + ( l + x) + ... + ( l + x + ... + x""^) 2 2 lim 2 " sin ^ 2".sin-^ l + x + x2+... + x " ' - ' 1 + x + x^ +... + x " ' 2" 2" t a i'.'Jir] I ' , - j j ••• m(m-l) n(n-l) = ' l + 2 + ... + ( m - l ) l + 2 + ... + ( n - l ) sma 1- sma 2" sma ,,. ., m m 2 sm— sm- 2" 2" m-1 n-1 m-n 2 2 2 III. B A I T A P T i ; L U Y E N C O D A P S O . jj,, , I , • „ %/cosax-Vcosbx ^, ,^ . cos — cosx B a i 7. T i n h L = l i m -^^ (a, b, c la cac so thifc khac khong) m , i 2 B a i 1. T i m g i d i han sau: C = Hm — ; "-»(' sin cx x-»o sin(tanx) -H» 2" B ^ i 3. T i n h g i d i han I = l i m x-»() lAvtdng dSn giai HrfdngdSngiai A ^ a a a a a a a A = cos —.cos — . . . c o s = COS .cos , n - lr . . . C O S - T - C O S — 2 2 2^^ 2^^ 1 ^ . a a a a a e" -1 ^^"'^ l n ( l + x^) /sm — cos 1= lim + -x .cos r...COS-T-COS— 2" 2" j ,n-l 2 ' x-+() x^ 2(l + V ^ ) Y x 2 ' l 4 , 2 sin 2" , 2 ,
  6. Luyfn giil di trudc thi DH 3 miin Bdc, Trung. Nam Todn hpc - NgiQ^n Van Tlidng C6ng ty TNHH MTVDWH Khang V toan xdl giao diem cua do thi y = f(x) vdi drfdng thing y = m (vuong gdc vd true tung va c^t true tung tai diem cd tung dp m). C A C B A I T O A N K H A O S A T H A M s6 4. SI/ tiep xuc cua hai dudng cong. • a. Dinh nghla: Hai ham so' f(x) va g(x) tiep xiic nhau tai diem M(xo; y,,) neu M I. T 6 M TAT LI THUYET: la mot diem chung cua chung va hai di/dng cong cd chung tie'p tuye'n vdi 1 . Cac bu6c khao sat si/ bien thien va ve do thi cua ham so nhau tai M . Bifdc 1. Tim tap xac djnh cua ham so b. Dinh It: Hai diTdng cong y = f(x), y = g(x) tiep xuc vdi nhau khi va chi khi he Btfdc 2. Xet sU' bien thien cua ham so ff(x) = g(x) phiTdng trinh an x sau day cd nghiem: \ - Tim gidi han tai v6 cifc va gidi han v6 cifc (neu c6) ciia ham so. Tim cic [f'(x) = g'(x) dircJng tiem can ciia do thi (neu c6). l> " (He phiTdng trinh nay la h? phiTdng trinh \ic dinh hoanh dp tiep diem cua - Lap bang bien thien cua ham so, bao gom: Tinh dao h^m cua ham so, xet haidirdng). ' - • dau dao ham, xet chieu bie'n thien va tim ciTc tri cua ham so (neu c6), dien II. C A C B A I T O A N M I N H HQA. . cac ket qua vao bang. ..^.^ ^ Bai 1. Cho ham so y = f(x) = - x ' + 2x^ + | x "^^ BMcJ. Ve d6 thi cua ham so. ^^ j - Ve cac du'dng tiem can cua do thi (neu c6). a. Khao sat h^m so. - Xac dinh mot so diem dac biet cua do thi, chang han tim diem uon, giao b. Tiep tuyen cua do thi (C) cua ham so tai goc tpa dp, cat (C) tai diem M . diem cua do thi vdti cac true toa dp (trong triTdng hdp do thi khong c^t cac Tinh tpa dp diem M . true toa dp hoSc vice tim tpa dp giao diem phtJfc tap thi bo qua phan nay),... c. Bien luan theo k so giao diem cua (C) va du'dng thing (d): y = kx. - Nhan xet ve do thj: Chi ra true va tam doi xuTng cua do thi (neu c6, khong n ; Ia Hi^ngdlngiai yeu cau chufng minh). a. Khao sat ham so vi£a>t, 1 2. Diem uon cua do thj: la diem U(X(); yo) ciia do thj sao cho lie'p tuye'n tai dd l.Tapxacdinh:D = R - di xuyen qua do thi, tiJc la ton tai mot khoang (a; b) chuTa diem Xo sao cho J/ 2. Sir bien thien: tren mot trong hai khoang (a; x,,) hoSc (X(i; b) tiep tuye'n cua do thi tai diem a. Gidi han: U nam phia tren do thi, con tren khoang kia tiep tuyen n^m phia du'di do thi. ^ o 2 5 Ta thU'dng suT dung ket qua sau day de tim diem uon cua do thi: lim y = lim -X +2x + - X - -00 X-»+00 X->-H» Neu ham so y = r(x) cd dao ham cap hai tren mot khoang chuTa diem Xo, f"(xo) = 0 va r'(x) doi dau khi x di qua diem Xo thi U(X(); f(X())) la mot diem lim y = lim (-x 3 +2x . ...2 + -5x ). = +oo X->-QC X->-
  7. Luy?n gidi dS IruOc kp thi DH 3 miin Bdc, Trung, Nam Todn hgc - NguySn Vdn Th6n^ g H a m so" nghich bien tren cac khoang —;+oo TrUdng h k = - thi(l)c6dung2nghiem. 1 mms^datcirctieutai x - - - y^=—i.''> a»arb m y i i . o u i l i l + N e u g(0) 9t 0 o k ;t - thi (1) c6 dung 3 nghiem o (C) va (d) c6 diing 3 3. D6^thi (Hoc sinh W ve) = ^ _^^y. ^ g,^,^ ^0^3^,}^ jj^H ^^J^^Q J| 3 , ,1)/, . ., + D i e m uo'n , -/.v i « ,r/-if>'t) ,, d i e m chung. ^ . n 2 / 2 46 B a i 2. Cho h a m so y = f(x) = - x ^ + 3x^ + 3(m^ - l ) x - 3m^ - 1 (1), m la tham so. y " = - 6 x + 4, y " = 0 o x = - . D i e m l -;— la d i e m uon. 3 , .3 2V J a. Khao sat s i f b i e n thien va ve do thi ham s 6 ' ( l ) k h i m = 1. hi b. T i m m de h a m so' (1) c6 cifc dai, ciTc tieu va cac d i e m cUc t r i cua do thi ham + x = 0=>y= — ^''^>''< l-^^''''" so (1) each deu goc toa do O. '2 46^ HrfdngdSngiai .,4,)(„f, . + NhSn x6t: D 6 thi nhan diem I l a m tam d o i xtfng. a. Khao sat h a m so (Hoc sinh tif giai). „ , . > 3'27 mi b. Ta c6 y ' = - 3 x ^ + 6x + 3(m^ - 1 ) ; ' b. PhiTdng trinh tiep tuyen (A) cua (C) tai O (0; 0) la: rx = l + m lv*i-.--K;vM.!:... • .Won y'=0« y = f ( 0 ) ( x - 0 ) + 0 = ^x SB" 0* [x = 1 - m D i e u k i e n de h a m so c6 ctfc dai, cifc tieu la y ' = 0 c6 2 nghipm phan biet PhiTdng trinh hoanh do giao d i e m ciia (A) v^ (C) IS: om?
  8. o 2m(xf, - 1)= y„ + XJ5 - 1 ; V m a. K h a o sdt v a v e do thi (C) c u a h a m so ( I ) . X„ =1 : • b T i m c a c d i e m thuoc (C) c6 loa do la nhiTng so n g u y e n . xf,-l =0 c T i m d i e m M e (C) d e long k h o a n g e a c h liT M d e n hai du'dng t i e m c a n c u a X() = - 1 • ( 1 ) Ub , E }\ V '= (O'k tr-.v; + yo+x^-l =0 (C) nho nhat. yo=l-Xo ;• ^ HU
  9. U^ngua ae miOc Ic m uii t mien oat;, irung, num ivunrm- lyguyen vun i tivng- Goi M (X(,; y„) e (C). Khi d6: III. C A C BAI T O A N T I ; L U Y f N c 6 H U 6 N G D A N d(M;d,)= |xo-l| Bai 1. Khao s^t sir bie'n thien vk ve do thi (C) cua hkm so y = f(x) = x^ - 3x^ + 1. '2x0-1 . 1 d(M;d2)= y o - 2 -2 1\i do, bien luan theo m so nghiem cua phiTdng trinh sau: -1 Xo Xo-1 , .v!' Khi d6, tong khoang c^ch \.\S diem M(X(); y o ) den hai di/cfng tiem can Ik -x^-x2+m+2=0 d = d(M;d,) + d(M,d2) ^ . . . ,1 HiMngdSngiai o^Ti, in, ? , Xo-1 + =2 • m < - 2 hokc m > —: Phi/Png trinh c6 mot nghiem. x„-l Xn -1 2 1 Xo=0 •dn,i„ = 2khi X Q - I • m = - 2 hokc m = —: PhiTPng trinh c6 hai nghiem. Xn -1 o ( x o - i r = 1 0 Xo-2 1 S m -J hod t. • xo = 0 = > y „ = 1 =>M3(0; 1) 2 • - 2 < m < —: 3 PhiTPng trinh c6 ba nghipm. f , , • u) ;.u ~ f ..' • Xo = 2 z:> y o = 3 =^ M4(2; 3) Bai 6. Cho h a m s o y = f(x) = B a i 2 . C h o h k m s o y = f(x) = x ' - m x + m - 1 (1) x2x+ 1 Aihly.-- a. Chtfng minh r^ng tie'p tuyen cua do thi hkm so (1) tai diem Xo = 0 c6 he so a. Khdo sdt sif bie'n thien va ve do thi (C) ciJa ham so' da cho. goc nho nhat. b. Tim toa do diem M e (C), bie't tie'p tuyen cua (C) tai M c^t hai true toa dp b. Vdi gia tri nko cua m do thi cija hkm so (1) tiep xuc vdi Ox. Khao skt vk ve Ox, Oy tai A, B va AOAB c6 di?n tich bkng - . _ do thi hkm so (1) gik tri tim diTpc cua m. 4 c. Xkc dinh m do thi hkm s6'(l) c^t true hoknh tai 3 diem phan bi^t. HUdng dSn giai ' Hi/dng d i n giai a. Khao sdt hkm so (Hoc sinh tiT giai). a. f ( X ) = 3x^ - m > 0 - m = f (0) '^^ ' b. Gpi M (xo; yo) 6 (C); y„ = f(xo) => hp so gdc tie'p tuyen tai diem x = 0 Ik nh6 nhat. ^^' * PhiTdng trinh tie'p tuyen2x^cua (C) tai M(xo; yo) la (d): y = f (xu)(x - XQ) + f(xo) b. m = 3; m = — ,41 li - y= -x + - 4 c. x' - mx + m - 1 = 0 o (x - l)(x^ + x + l - m ) = 0=> (Xo+1)' (Xo+1)' r m > 04 2 'i'"' ' . J i i f P ' M ' k - Toa dp giao diem A cua (d) vk Ox Ik: A( -Xo ; 0) Bai 3. ( 2x2 a. Khio sat si/ bie'n thien vk ve do thi (C): y = f(x) = x' + x - 1. Tpa dp giao diem B ciia (d) va Oy la: B 0; ^ b. Gpi Xo Ik nghipm cua phiTPng trinh f(x) = 0. Chtfng minh r^ng: De thay AOAB vuong tai O S^QAB =^0A.0B 2Xo - X(, - 1 < 0 M (trr 1-1^2 H i ^ n g d i n giai 2x2 2xo + X() + 1 = 0 Xo=-- Xn. 4x^-(Xo+1)2=0: f(x) lien tuc tren 4 2 "(xo+1)^ 2 x 2 - X o - l = o' Xo=l Ket luan: M, ;M2(1;1) if ( n = 17 . r ViEN T I W H B I N H THUAN V 2
  10. Luyfn giii di trade kp thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyln Van ThOng b Phu'dng irinh hoanh do giao diem f(x) CO n g h i e m X(, 2 (2x,l-6xo)x--x^3xi,+- =^ - 3 x + - •(xo- 1) x„ + — < 0 => 2 x f ) - X ( , - 1 < 0 (dpcm) o(x-xo)'(x^+2x„x +3x^6) =0 x/r .-:,,„• B a i 4 . C h o ( C J : y = f(x) = mx' + ( m - l ) x ^ + 1 - 2m ' Bai 7. Cho ham so y = r(x) = p a. Tim m dc ham so' co mot diem cifc tri. b. Viet phiTcfng trinh tiep tuycn cua f C , ^ di qua g6'c toa do. >f{ £ > tu » a. Khao sat sir bien thicn va ve do thi (C) cua ham so. . 2) b Tim m de di/c^ng thang (d): y = 2x + m c^t (C) tai 2 diem phan bi^t A va B. tim tap hdp trung diem I cua AS. t. m + A« j 8- ^ (£)^j '"" ' a. m < 0 hoac m > 1 Hrfdng dSn giai '(.m.utll b. (d,): y = 0; (d:): y = ^ x ; (d,): "'^ b —=- = 2x + m o 4. ' a . Khao sat va ve do thi (C) cua ham so'. TiT do, bien luan theo a so' nghiem cua Tap hdp trung diem I cua AB la phan diTdng thang 2x + y - 4 = 0, gom phi/dng trinh x"* - 2x^ + 11 - a = 0. ..x- rn + 4 m + 4'\ . ik>M..4 b. Tim m de parabol (P): y = mx^ - 3(m^0) tiep xuc vdi (C). ,, nhffngdiemM ; vdim4. , :,,[•••. v 4 2 y . — - Htfdng d§n giai .^j,,, 2x — 1 a. • a < 10 : Phi/cfng trinh v6 nghiem Bai 8. Khao sat va ve do thi ham so y = l(x) = . Tilf do, tim m de phiTdng x +2 • a = 10 : Phu'dng trinh c6 1 nghiem. _ ^ r , . 2sinx-l . , , - ... '* 'sil »tn -.i • • 10 < a < 11 : Phu'dng trinh c6 4 nghiem. j ^'g; trinh = 2m - 1 c6 dung 2 nghiem x x; sin X + 2 • a = 11 : PhiTcfng trinh c6 3 nghiem. HiAJngdSngiai al =.(x)t mil • a > 11 : Phifdng trinh c6 2 nghiem. * Khao sat va ve do thj (Hoc sinh tiT lam) b. (P) tiep xuc vdi (C) khi va chi khi hoanh do tiep diem la nghiem cua he * Dat t = sin x; x e [0; 71J ^ t e [0; IJ n « (,., •^rmd^ fa; ; phiTcfng Irinh PhiTdng trinh trd thanh: = 2m - 1 (1) " ' x^-2x^ + l = m x 2 - 3 4 •m = -2 ft) Dieu kien de phiTdng trinh c6 dung 2 nghiem x e [0; n] la phtfdng trinh (1) co 4x -4x=2mx 7^ dung 1 nghiem t e [0; 1). Bai6.Cho(C):y = f(x)= - x ' * - 3 x ^ + - . Khao sat ham so g(t) = = ^ iren [0; 1) => - < g(t) -
  11. I ¥unf^, t^urri i utin n\n, - ii^ujfcri run i riurig COngtyTNHHMl ^ L v vii KHung vict Hifdng d i n giai a) De ddn giSn, ta dat A = 3a^ - 1, B = -(b^ + 1), C = 3c^ D = 4d, h^m so da H^m so nghjch bien tren moi khoang (-oo; - 2 ) , (0; +oo) va dong bien tren cho chinh m y = Ax^ + Bx^ + Cx + D khoang (-2; 0). Ta CO y' = 3Ax^ + 2Bx + C y' = 0 « • 3Ax^ + 2Bx + C = 0 Ham so dat ciTc dai tai x = 0 va dat ciTc tieu lai x = - 2 . PhU"dng irinh nay c6 hai nghicm phan bi^t la x = 1, x = 2 nen ta c6 h? sau • Bang bie'n thien '• I i . ' .• f3A + 2B + C - 0 ^(^)i=^= Y o ^ m i f i i , . . ! r , i « a —00 -2 0 +C0 0 _12A + 4B + C = 0 , „x ^ +00 g _ ry(l) = _7 [ A + B + C = -7 rniT / Ta cung c6 0 -00 y(2) = - 8 ' ^ l8A + 4B + 2C + D = -8 f> ' • Ta c6 y" = -6x - 6, y" = 0 TCf cdc dieu kien n^y, ta tim diTdc A = 2, B = - 9 , C = 12, D = -12 ta tinh o X = - 1 nen diem uon cua diTcIc cdc gia tri tiTcfng tfng la a = ±1, b = 2, c = ±2, d = -3. - ^ do thj la (-1;2) V a y M = a^ + b^ + c^ + d^= 1^ + 2.2^ + 3^= 18 x ^ i Do thi cii true lung tai diem b) Phi/dng trinh hoanh dp giao diem cua hai do thj Ik (0; 4) va cat true hoanh tai x" + 2 m V + 1 = X + 1 o x(x' + 2m^x - 1) = 0 .-j.,, diem(l;0),(-2;()). De thay phifdng trinh nay co nghiem x = 0, ta can chifng minh phifdng trinh Do thi cua ham so: t c6n lai c6, nghipm duy nha't khac 0. b) Ham so da cho nghich bien That vay, xet ham so f(x) = x ' + 2mx^ - 1. tren khoang (0; +co) khi va Ta c6 f (x) = 3x^ + 2m^ > 0 nen ham so nay dong bien tren tap so thi/c. chikhi ^ , X Hdnnffa lim f ( x ) = lim (x"^ + 2m^x - 1 ) =-oo y' = -3x^ - 6x + m < 0. Vx > 0 Hinh 7. Do thi ham soy = - J C ' - 3 X ^ - ^ 4 Va lim f ( x ) = lim ( x ' ' + 2 m ^ x - l ) = +oo » 3x^ + 6x > m, Vx > 0 Ta CO bang bien thien cua ham so g(x) = 3x^ + 6x tren (0; +oo). Nen phi/dng trinh f(x) = 0 luon c6 nghi$m. +00 0 Suy ra phi/dng trinh f(x) = 0 c6 dung mot nghiem nay cung khdc 0 do +00 f(0) = - l . T a c 6 d p c m . i_ Bai 2. Cho ham so y = -x^ - 3x^ + mx + 4 vdi m 1^ tham so thifc Tir do, ta difdc dieu kiOn cua m la m < 0. a) Khdo sat sir bien thien va ve do thi do thi hhm so khi m = 0. Bai 3. Cho ha m so y = f(x) = Sx'* - 9 x ^ + 1 r ; . • b) Tim tat ca cac gid tri cua tham so m de h^m so da cho nghich bien tren a) Khao sat sir bien thien va vc do thi ham so tren. - i 1 '' - • (0; +00). b) Dira vao do thi tren, hay bicn luan iheo m so nghicm cua phiri^ng trinh Ht/dng dSn giai ^ 8cos'x - 9 c W x + m = 0 vdi X e [0; 71]. a) Vdi m = 0, ta c6 h^m so y = - x ^ - 3x^ + 4. HU-KC x->-a! • Chieu bia'n thien: Ta c6 y' = -3x^ - 6x, * Chieu bien ihicn: y' = 0 o -3x^ - 6x = 0 X = 0, x = - 2 . y 3 Ta c6 y' = 32x' - 18x. y' = 0 x = 0, x = ± - . ii 0 m c:^ i
  12. COng ty TNHH MTV DWH Khang Vi?t Li^n giii di tniOc Aj> thi DH 3 miin Bdc, Trung, Nam Todn hQc - NguySn Van Thdng Ncu 0 < 1 - m < 1 o 0 < m < 1 Ihi phiTdng Irinh c6 dung hai n g h i e m . 3 H a m so dong b i c n tren — ;+tx3 A',, < 1- m < 0 o 1< m < ^ Ihi phifdng U-inh c6 d u n g b o n n g h i e m . 4 Neu 32 32 Va nghich bi6'n tren ( -co; 3^ 0; — Neu I - m = m = | i ihi phi/dng trinh c6 dung h a i n g h i e m . I 4; 4j 32 32 ( 3. 49 ^ r3. 49^ B a i 4. Cho h a m s 6 > = f(x) = m x ' + 3 m x ' - (m - l ) x - 1 v d i m la tham so'. ' J H a m so dat ciTc dai tai (0; 1) va dat ciTc tieu tai . 4 ' 32 J . 4 ' " 32 a) Khao sat sir bi en thien va ve do thi cua ham so khi m = 1 X ,X b) Xac djnh taft ca cac gia t r i m dcf ham scTy = l ( x ) khong c6 cifc tri. -00 0 +00 4 4 Hif(}ng d§n g i a i ' y' - 0 + 0 0 + a) V d i m = 1, ta CO T a p x^c dinh D = M . h a m so y = x-* + 3 x ^ - 1 . ^x,^ .i.'-yr ,.. , 5}^ , , y +00 ^ 1 ^ +CO ^ 49 49 ^ G i d i han cua h a m so: > / 32 32 lim y = l i m ( x - ^ + 3 x ^ - 1 ) = - . , va j i m y = l i m ^ ( x ^ + 3 x 2 - 1 ) = + « ) . X->-OD X->-00 x->+oo Ta CO y " = 9 6 x ' - 18, y " = 0 o x = ± — , y = - — n e n hai d i e m uon cua do * Sy bien thien ciia ham so: 2 Ta CO y' = 3x + 6x, y' = 0 o x = 0, x = - 2 . ^^/3 13^ H a m so nay dong bi en tren cac khoang (0; +00), ( - c o ; - 2 ) va nghjch bien tren thi ham so nay la 4 ' 32 4 32 (-2; 0). * Bang b i e n thien: Do Ihi cua h ^ m so c^t true tung tai d i e m (0; 1) va c^t true hoanh tai bon 1 1 d i e m phan biet la ( - 1 ; 0), ( 1 ; 0)J ;0 n 4 .2N/2 , •3 • D o t h i ham so 'y y = 1 - m Ta CO y " = 6x + 6, y " = 0 o X =-1 b) Dat I = cosx, X e [(); n ] , ta c6 (d) n e n d i e m u o n c u a do thi la ( - 1 ; 1) t e [ - l ; 1] va ro rang m o i gia trj Do thi h a m so c a t true tung tai d i e m + cua t (t e [ - l ; 1 ] ) cung cho ta duy "K 1/ ( - 1 ; 0) v a c a t true h o a n h tai ba d i e m ' nha't mot gia tri cua x. phiTdng 3 / \ "V I p h a n biet. \ X It trinh da cho ti/dng diTdng v d i 1 4 / >. i * Do thi h a m so: ' 1 1 f(t) = St' - 91^ + 1 = 1 - m -2 -1 1 2 x /o h) Ta x e l c a c trU'ctng hdp sa D a y chinh la phifdng trinh hoanh do giao d i e m cua hai do thi - K h i m = 0 thi y = X - 1 ''^ y = 8x''-9x'+l,y= 1-m. ~ 32 nen h a m so k h o n g CO cifc trj. -2- Di/a vao do thj, ta thay r^ng - V d i m ^ 0 thi y' = 3 m x ' + 6mx - m + 1. 49 Him so n a y k h o n g c6 cifc trj khi v4 Ncu l - m > l v l - m < - 32 Htnh 8. Do thi ham soy = 8x^-9x^ + 1^ c h l k h i phiTdng Irinh y' = 0 k h o n g c6 81 n g h i e m h o S c c6 n g h i e m k e p , tiirc la o m < 0 V ni > — ihi phiTdng trinh vo nghiem. A ' < 0 « . 9m^ + 3 m ( m - 1 ) = 12m^ - 3m < 0 0 < m < ^. Neu 1- m = 1o m = 0 ihi phiTdng trinh co diing mot n g h i e m .
  13. Luy(n giii di trade kj> thi DH 3 miSn Bdc, Trung, Nam ToOn hoc - NguySn Van ThOng Ceng ty TNHH MTV DWH Khang Vi(t VSy dieu kien can nm la 0 < m < - . Phildng tiinh nay c6 ba nghi^m phan bi^t Ik 1 - 2>/3 , 1 , 1 + 2%/3 thoa man de bai. ^ 't> 0 > m > • • Vay gia tri can tim cua m la m = 11. B&i5.Chohamsoy = - 2 x ' + 6 x ^ - 5 c 6 d 6 l h i ( C ) Bai 7. Cho hkm so y = f(x) = x ' - 2x^ c6 do thj (C). f"" ' '
  14. Luy^n gidi di truOc thi DH 3 mijn Bdc. Trung. Nam Todn hoc - NguySn Van Thdne Ceng ty TNHH MTVDWH Khang Vi?, ICA = ke « 4 a ' - 4a = 4 b ' - 4b c:>(a - b ) ( a ' + ab + b ' - 1) = 0 Do A va B phan bict n c n a * b, suy ra a ' + ab + b ' = 1 (2m^-l)x-m _^ [(2m-l)x-m^ =x^-x ](x-m)^=() M 3 l k h i c , hai tiep tuyen cija (C) t a i A va B irung nhau k h i va chi k h i ^ . .j. x - l = ±(m-l) x - l = ±(m-l) a- + ab + b^ = l,a ^ b ^ a^ + ab + b^ = l,a ;t b ( h% ; ^j,. ^)|,;f>{ (m-ir ^ f ( a ) - a f ' ( a ) = l"(b) - bl"'(b) [-3a^ + 2a^ = -Bb"* + 2b^ \ •
  15. Luy(n gidi dS trudc thi DH 3 miin Bdc. Trung, Nam Todn hoc - Nguyin Van Thdng Nen ham so da cho nghich bien tren tiifng khoang xdc dinh. = (xi - x j r Ham so da cho khong c6 cifc tn. , -v^ » v> < - i'- a . i . j V 4xfx^ J , a c dir&ng tiem can: Ta c6 pB t ar 6. r > u v M . fr. < ; T ' X ^ ' X ^ Theo dinh li Vi-et cho phuTdng trinh (*) thi x, + X j = -^^3_i2 ^ XjXj = - 1 - -00, lim y = lim = +00 Urn y = hm 4(x-3) P 4(x-3)J 4(q-ir ^ 1 Dod6 10 = ((x, +X2)^-4x1X2) 1+ +4 4xfx2^ lim = lim 4(x-3)j D6 thi CO X = 3 la ti?m can dilng va y = j la ti?m can ngang, tarn doi xuTng 1 + . 1 > X > !• O S2:jx m Bii-J,nh ill :m ,1 > A r • Theo gia ihiet + = 1o = 1 _ q^, ihay vao ding thuTc iren. ta diTdc la I 3 ; - . V 4j Vai : 10 = ( ( q - l ) ^ + l - q ^ ) 1 + - * Bang bien thien 1; +00 X -00 10(l-q^) = ( 2 - 2 q ) ( l - q ^ + 4 ) o q - % 4 q ^ - 5 q = 0 y' - - Giai phifdng trinh nay, ta thu diTdc nghiem thoa man de bai la q = 0. Thuf lai 1 +00 y ta tha'y thoa. Vay cac gia tri can tim la (p; q) = (1; 0), (p; q) = ( - 1 ; 0). 4 1 —00 4 b) Ta CO y = x + m + x +1 * Do thi ham so c^t true tung va true hoinh tai (0; 0). (ve h\nh) 1 b) Goi M(X(,; y„) la mot diem thuoc do thi. Khi do. tiep tuyen cua do thi tai M \k y'=l-- , y'= 0 o (X + 1 ) - = 1 o X = 0 , X = 2. (x + i y K.. -3x . 4xf)-9X() -3 _, -(X-X(,) + = y Do do, hai diem cifc trj cua do thi chinh la (0; m + 1), (-2; m - 3) nen khoang y= 4(xo-3) 4(Xo-3)' 4(xo-3)' 4(Xo-3) each giffa hai diem nay la d = V(0 +2)^ + (m + 1 - m + 3)^ - 2 y / 5 khong doi. fo.4xf,-9x„ Giao diem cua tiep tuyen vdi true tung c6 tpa dp la , giao Ta CO diem phai chiJng minh. | p + »:} +^x>yC.--(l ^ ^xVo + xS) '4(Xo-3)^j c) Hoc sinh tirkhao sat f4xS-9x„. Bai 4. Cho hamso y = c6 do thi (C). diem cua tiep tuyen vdi true ho^nh c6 tpa dp l i ;0 4(x-3) ft a) Khao sdt va vc do thi cua ham so da cho. „ ,. , , v . Do d6, tir dieu ki^n de b^i, ta c6 '"^ b) Tim tpa do diem M thupc (C) sao cho tic'p tuycn cua (C) tai M c^t hai true 1 (4xf)-9x„)^_3 2 .2_a.. _3)2 j lit. tpa do Ox, Oy Ian liftJt tai hai diem A, B v& dien tich tam giac OAB \h -. 8 c) Tim nhurng diem u-en (C) co khoting each den true hoknh gap 3 Ian den true tung. Gidi phirpng trinh nky, ta thu diTpc 3 nghicm 1^ x„ = ^ , x,, = ^ ( l ± ^ ) I 2 4 Hif(}ng dSn giai a) Tap xac dinh D = ( - 0 0 ; 3) u ( 3 ; + 0 0 ) . . , . ,, , ,^ Ttr d6 ta xac djnh diTpc cic diem thoa man de b ^ i . * Sirbienthiencuahamso: , B i i 5. Cho h^m so y = c6 do thi (C). , Ta CO y' = • -3
  16. o o n g ly iisnti M I V uvVH'Khang Vii b) V d i m o i d i e m M ba't k i thuoc do thi (C), t i m gia tri nho nha't cua tong khoang T o n g khoang each tiT M t d i hai diTcJng t i e m can ciia (C) la each tCr M den hai true toa do. , >, j , n ;, ^ t,tij HU (.,• t,\m-' 1 c) T i m nhOrng d i e m tren (C) C O toa do la cap so nguycn. ,•>',,, ; '.],.,.'} 2xn- 2x„-l + X,) - 1 1 !' HMng dSn giai \ x» - I d = |xo-l N/5 Xo-1 til a) Hoc sinh tU" khao sat. ' ' * (*,,**' Ta c a n tim gia trj nho nha't c u a b i e u thuTc n a y iJng v d i m o i gia tri x,, ^ 1. b) Theo de bai, ta can t i m gia t r i nho nha't cua bieu thiJc sau \ 1 2 ! x-1 f':'~; mil \ 3 vf;"'!'?'- Taco d>2 Xa -1 A = + 2x + 3 x„-l U(j ',,1' 5 " Ta tha'y rang v d i x = 1 thi A = 1 nen ta chi can x6t cac gia trj cua x khac i 2 Suy ra gia u-i nho nha't do c h i n h la d = - ^ ^ , dat diTdc k h i sao cho A < 1, turc la chi can xet |x| < 1 o - 1 < x < 1 . K h i do, do 2x + 3 > 0 nen ta chi can xet 2 tru'dng hdp sau: x„-l X n = 1 ± x„-l -X -2x^-4x + l Ne'u-1 < x < O t h i A = -x + = f(x). 2x + 3 2x + 3 1+ 1 ;1+ 2 V a y d nho nhal k h i M = Ta C O l ' ( x ) - — < Q nen day la ham nghich bien, suy ra (2x + 3r Hoac M = l-4=;l-i-^ gnu J V f(x)>lXO)= i. (x-l)-^+a + l B a i 7. Cho ham so y = X Ne'uO
  17. J ) T i m tham so Ihifc m dc tpa dp cifc dai va ciTc l i e u cua ( C m ) n a m ve hai phia ( x ^ + x + l ) - ( x + a)(2x + l ) _ x^ + 2ax + a - 1 b) Ta CO y' = :i 5 — 5 5— cua dirdng t h i n g 3x + 4y - 7 = 0. - (x^+x + i r f (x^+x + i r Hufdng d§n g i a i ii; v , ' • 2(xU3ax2+3(a-l)x-l / ^ , a) Hoc sinh tif g i a i m :uiT> •:i)n (» i 39 A' = ( m - 1)^ + 3 ( m + 3) = m^ + m + 10 = >0 x' + 3ax' + 3 ( a - l ) x - 1 = 0 (•) i j , " ^ ^ ',1 . , , ^ ~ -. j; jj 1 nan luon cd 2 n g h i e m phan biet. Suy ra v d i m p i m , h a m so c6 ciTc d a i , ci/c C O dung ba nghiem phan biet. , cv' i l - ,,/i'lV D a t f ( x ) = x ' + 3ax^ + 3 ( a - l ) x - l , x € R. ticu. Thi;c h i c n phep chia y cho y', ta cd. m-1 Ta C O f(0) = - 1 < 0, f ( - l ) = 1 > 0, l i m f ( x ) = - o o , l i m f ( x ) = +oo dong y = x ' + ( m - l ) x ' - ( m + 3 ) x - 1 = ( 3 x ' + 2 ( m - l ) x - ( m + 3)) — + • 3 thdi h^m so nay l i e n tuc tren tap so thifc nen phiTdng trinh f ( x ) = 0 c6 ba nghiem phan biet thuoc cac khoang ( - o o ; - 1 ) , ( - 1 ; 0), (0; + o o ) . m^+2m-12 2(m-l)^ 2 ( m + 3) Do do, phi/dng trinh (•) c6 ba nghi?m phSn biet hay do t h i da cho c6 ba d i e m x +- uon. 2(m-l)^ 2 ( m + 3). m^+2m-12 G'd sur ho^nh do cua mot trong cac d i e m uon cua do t h i h ^ m so da cho 1^ x„ Do y c D = X c D + - va day cung la nghiem cua phiTcfng trinh (*) hay X Q + 3axo + 3(a - 1)X(, - 1 = 0 ; 9 3 :ntfd 1' khi d6, tung do trfdng uTng cua d i e m n ^ y chinh 1^ yo = —5-^ . Ta se t i m 2(m - 1 ) ^ 2 ( m + 3) m^ + 2 m - 1 2 XcT + Xo + Xo +1 ycT = mot quan he tuyen tinh giffa Xo, y o . Suy ra cac d i e m cifc dai va ciTc tieu n^m tren di/dng t h i n g cd phifdng trinh Tir dieu k i ^ n cua Xo, ta thafy r^ng ^ 2(m-l)^ 2(m + 3 ) 1 ^ ^ m ^ + 2 m - 1 2 X Q + 3axf, + 3 a x „ + 3a - 1 = 3 x „ + 3a o (x,) + 3a - l)(xo + x,, +1) = 3 ( X Q + a ) . ' y = 2 :d S „ y ra y„ - + x„ +1) ^ XQ + 3a - 1 ^ va do chinh la phUdng trinh diTdng t h i n g di qua hai d i e m ciTc tri. 3(Xo + x„ +1) .3 c) N c u d i e m A cd tpa dp lii (x; y) thi d i e m d o i xtfug cua A qua diTdng t h i n g Do d6, cdc d i e m uon cua do thi cung thoa man quan h$ tuyen tinh nhif tren, y = X cd tpa dp la (x; y). V i the y c u cau cua bai toan tiTdng diTdng v d i viec ttfc Ik chung t h i n g hang. Du'dng t h i n g di qua cdc d i e m uon tifdng tfng chinh x+ 3a-l tim nghiem nguyen (x; y) v d i x^y cua he phi/dng trinh I la y = . S . + r , r [x = y " ^ - 4 y - l MQT S6 DANG TONG H0P TOAN LIEN QUAN DEN ^ Trir hai phi/dng irinh ve thco ve, sau do chia hai ve' cho x - y ;>t 0, ta dU'dc + xy + y^ = 3. DO TH| HAM S6 NANG CAP ' X' Dc dang t i m dU'dc nghiem nguyen v d i x^y cua phu'dng trinh x ' + xy + y^ = 3 Bk\. Cho ( C „ ) c6 phifdng trinh y = x ' + ( m - l ) x ' - ( m + 3)x - I . ia(2;-l),(-l;2),(-2; l),(l;-2). "^'^ a) Khao sdt ve do t h i (C) ciia hkm so k h i m = 1. Thur lai vko he, ta nhan bp nghiem ( 2 ; - 1 ) . ( - 1 ; 2 ) . ' ' ' b) Chtfng m i n h r^ng v d i m p i m, hkm so' c6 ci/c d^ai, ci/c t i € u . Vie't phifdng trinh V a y la t i m dirpc cap d i e m nguyen duy nha't d o i xuTng v d i nhau qua di/dng dirdng t h i n g d i qua c i c d i e m cifc d a i ciJc tieu cua do t h i . t h i n g y = X va khong nam tren di/dng t h i n g do la (2; - 1 ) , ( - 1 ; 2). , ...... c) T i m nhffng cap d i l m nguyen tren (C) do'i xtfng v d i nhau qua diTdng t h i n g y = X v£l khong nKm t r e n di/dng t h i n g d6.
  18. Lu^n gUu di trUOC thi DH 3 m,.'n / U /' \.,:m / >i h \ . , • 7 , ',,;/( Thdng CdngtyTNHH MTVl)\> / / . , Vi(t_ -x^ + mx - c6 diing hai nghiem phan bipt (an so la m). B a i 2. Cho ho diTcfng cong (C J : y = x-m Viet lai (1) diTdi dang m^ - (x,, + y„)m + X(,(x,i + y„) = 0, ta suy ra tat ca nhuTng a) Khao sat ve do thj ( C ) cua ham so khi m = 1. diem (xo; yo) thoa man yeu cau bai toan la nhCTng diem c6 toa do thoa man b) Xac dinh m de ham so c6 ciTc dai, ci/c tieu. Vie't phiTctng trinh diTdng thing di (x,) + y())(yo - 3x()) > 0. qua cac diem ci/c dai va cifc tieu cua do thj ham so'. Bai 3. Cho ham so y = x ' - 3(m + l)x^ + 2(m^ + 4m + l)x - 4m(m + 1). ( C J c) Tim cac diem trong mat phlng sao cho c6 dung hai di/cfng cua ho ( C J di qua. a) ChiJng minh rang ( C ^ ) luon di qua mot diem co djnh khi m thay doi. H i f d n g dSn g i a i b) Tim m sao cho ( C m ) c i t true hoanh tai 3 diem phan biet. —x^ + X — 1 • 1 c) Khao sat ve ve do thj ham so' khi m = 1 a) Khi m = 1, ham so trd thanh y = -x - x-1 ' x-1 H i M n g dSn g i a i * ^ ^ x (>: nM si ui il / f Mien xac d j n h D = R \ { 1 } . a) Gia suTdiem co djnh la (x„; y„). Khi do ta c6 f"^' tJ'' f^ 1 x(2-x) '^^te U Taco y = - 1 + = —^ yo = - 3(m + l)xf) + 2(m^ + 4m + l)x„ - 4m(m + 1) diing vdi moi m. Vict dang thuTc tren nhu'da thiJc theo m, ta difdc y' = 0 k h i x = 0 h o a c x = 2. ^, j (2X() - 4)m^ + (-3 xf, + 8x0 - 4)m + xf, - 3XQ + 2xo - y,, = 0 vdi mpi m. H^m so tang tren (0; 1) va (1; 2), giam tren (-00; 0) va (2; +00). Ham so dat ciTc tieu bing 1 khi x = 0 v^ dat ciTc dai bllng -3 khi x = 2. '2xo-4 = 0 ' 1 1 Dieu nay xay ra khi va chi khi -3xf) + 8x„ - 4 = 0 • Gidi han: Um y = lim -X -- = +00; lim y = lim -X - - -00: X->-oo X-+-00 x-lj X-++00 X->+00 x-lj xf,-3xf,+2xo-yo =0 lim y = lim -X - = -00; lim y = lim -X - = +00 Tur day giai ra dU'pc Xo = 2, yo = 0. x->l-' x-1 x-*i~ x->r x-1 Vay ( C m ) luon di qua diem M(2; 0) CO djnh. niiid H^m so CO tiem can xien y = - x va ti^m can dtfng x = 1. t») ( C m ) c i t true hoanh tai 3 diem phan biet khi va chi khi phtfdng trinh Bang bie'n thien x ' - 3 ( m + l ) x ^ + 2(m^ + 4 m + l ) x - 4 m ( m + 1 ) = 0 (1) —00 0 1 +00 CO 3 nghiem phan biet. TiT cau a) ta thay ring x = 2 la nghiem cua phiTdng 0 0 trinh tren. Nhd do, bie'n doi da thufc ve trai, ta diTdc phiTdng trinh tiTdng diTdng +00- r+00 (x - 2)(x^ - (3m + l)x + 2(m^ + m)) = 0 -00 -00 Tilf day ta thay r i n g (1) co 3 nghiem phan biet khi va chi khi phi/dng trinh (2) D 6 thi ham so: (hoc sinh tif ve Hnh) sau day CO 2 nghiem phan bietkhac 2: ^, 2 2 n fn»t - (3m + l)x + 2(m^ + m) = 0 (2) b) Ham so' y = ^ " ^ ^ - m ^, ^^^^ ^ . ^ x ( 2 m - x ) x-m (x-m^ |A = (3m + l ) ^ - 8 ( m ^ + m ) > 0 Dieu nay xay ra khi va chi khi 2 so y = -X + mx - m l 2 ^ - ( 3 m + l).2 + 2 ( m 2 + m ) ^ 0 c6 ciTc dai v^ ciTc tieu khi vS chi khi m^O. x-m Giaira ta diTdc m 1. Khi 66 hai diem cifc tri c6 Ipa dp ti/dng ufng la (0; m) va (2m, -3m). Suy ra Vay vdi m ; t 1 thi ( C n , ) c i t true hoSnh tai 3 d i l m phan bi$t. dudng thing di qua hai diem cifc tri c6 phiTdng trinh l i y = m - 2x. c) Gia sur (x,), y„) la mot diem trong mSt phing ma c6 dung hai diTdng cong ( C J B^i 4. Cho ham so y = — . ( C ) x-1 di qua. Khi do phifdng trinh y„ = '"^o ~ "^^ 3) Khdo sdt sir bie'n thien v^ ve do thi (C). (1) Xo-m ^) Tim hai diem A, B thupc (C) v^ doi xtfng nhau qua difdng thing y = x - 1-
  19. Luyfn giat ae truac Kytnttitijmi g Hifdng dSn giai ' (x-ir a) Hoc sinh tiT khao sal. !. if, i . f \ ^.r t>i|V pai 6. Cho ham so y = b) N e u M ( x , y) M ' ( x ' , y') 1^ hai d i e m doi xtfng v d i nhau qua du"dng thSng a) Khao sat sir b i c n thien va ve do thi ham so da cho. y = X - 1 Ihi ta C O . ,„., ,,„,,,,,.,, v=,. • (x-1)^ b) B i e n luan theo m so nghiem cua phi/dng trinh = m. x +2 i) - — - = - 1 (do dirdng th^ng M M ' vuong goc v d i diTdng t h i n g y = x - 1). x'-x Hifdng d§n giai , V V + Y ' X + X' ii) Trung d i e m cua M M ' nSm tren y = x - 1, suy ra ^ =— 1 a) Hoc sinh t i f g i a i . b) + m < 0: phiTdng trinh v6 nghipm Tir i ) va i i ) ta suy ra x ' = y + 1, y ' = X - 1. Gia suf M va M ' la hai d i e m thupc (C) doi xufng v d i nhau qua diTdng thang + m = 0: phiTdng trinh co 1 nghiem duy nhat; + 0 < m < 12: phi/dng trinh co 2 nghiem; y= -hav f ','•'•1: 'l:'. + m = 12: phiTdng trinh c6 3 nghiem; y = X - 1 thi ta C O + m > 12: phiTdng Irinh co 4 nghiem. , . x'^ (y + l)2 x - l = y ' = — — = ••' x'-l y + 1-1 Sd liTdc each g i a i : TiT do thi G d cau a), ta suy ra do thj G' cua ham so G i a i he nay ta diTdc c5p d i e m triing nhau M = M ' c6 toa dp 1 1 va cap v = ^^~'^ nhir sau: V d i x > - 2 , G' chinh la G. V d i x < - 2 , G' la anh doi x + 2 U 2) ••••X' , xiJng cua G qua true hoanh. diem M . D o chinh la nhi?ng d i e m A , B 72' ' V2J' ^ I sflj TCr do thi G' ta cd ket qua bien luan neu tren. can t i m . Bai 7. T i m m dc phi/dng trinh sau cd 4 nghiem phan biet: I > ^ • ^ / ^ l . 1.:. 2x^+(6-m)x + 4 - 3 x - l = mx-m Bai 5. Cho h ^ m so y = ^ mx + 2 Hi/(tng d§n giai a) Chtfng minh r^ng v d i m o i gi4 trj cua m , do thi ham so luon di qua mot diem Dap an. 1 < m < 6N/3-9. C O dinh duy nha't. Xic dinh tpa dp cua d i e m do. Cdch I. Chu y phiTdng trinh luon cd nghiem x = 1 v d i m p i m . Khiio sat ham so b) Khao sat va ve do thi cua ham s6' k h i m = 5. Hifctng dan giai -3x -1 y =- (x ;^ 1). Ta can t i m m sao cho difdng thang y = m cat do Gia sur (x,,; y,,) la d i e m co djnh ciia do thj ham so. x-1 L'u- 1 ' . ' 2x,^)+(6-m)x„+4 ^. thi tai 3 d i e m phan biet. K h i do ta C O y;, = — — ^ '—il vdi moi m. Cdch 2. Del thj ham so' y = 4|x|-^ - 3 | x | - 1 g6m hai nhanh, nhanh 1 la d6 thi ham mx,, + 2 so' y = - 4 x ' + 3x - 1 v d i X < 0 va nhanh 2 la d6 thj ham so' y = 4x- - 3x - 1 Suy ra mXd.yi, + 2y„ = 2 xf, + (6 - m ) X ( , + 4 v d i m p i m. vdi X > 0. T r o n g khi do y = mx - m la diTdng thdng quay quanh d i e m A ( l ; 0). V i e t phirong Irinh tren diTdi dang nhj thiJc iheo m , ta diTdc Hay vie'l phi/dng trinh lie'p tuye'n ke tiT A den nhanh 1 cua d6 thj ham so'. (xoyo + x,i)m + 2y() - 2 xf, - 6x1, - 4 = 0 v d i m p i x Suy ra Xoy„ + x„ = 0 vii 2y„ - 2 x,^, - fix,, - 4 = 0. h Jo. G i a i he nay ta diTdc nghicm duy nhal x„ = 0, y„ = 2. Tit do suy ra do thi ham so luon di qua d i c n i co djnh duy nha't c6 tpa dp (0; 2).
  20. BAI TAP T O N G H 0 P L U Y g N THI DAI H Q C V A L U Y g N H Q C SINH G I O I Phi/dng trinh hoanh do giao diem cua (C) va tiep tuyen co dang B a i 1.Cho ham so y = -x^ + 3x^ - 2. (C) ' '^ (3x^ - 3 ) ( x - X A ) + yA = x ' - 3 x + 2 a) Khao sat sir bic'n thien va ve do thj (C). Tif do vc do thj (C) y = + 3x^-2 Thay yA = - 3 X A +2 vao phUOng trinh, ta di/dc b) Tim tren (C) nhffng diem ma qua do chi ke diTdc mot ticp tuyen vdi (C). (3x^ - 3)(x - x A ) + X ' A - 3 X A = x-^ - 3x < : > ( X - X A ) ^ ( X + 2 X A ) = 0. H\i6ng dSn giai Nhi/ vay tie'p tuyc'n cua (C) tai A c^t (C) tai diem co hoanh dp XA (chinh la a) Hoc sinh tiT khao sat. A) va diem co hoanh dp - 2 X A (la diem A'), tufc la X A = - 2 X A b) Gia sur M(x„; y,,) la mot diem tren (C). Ta giai bai loan vict phiTOng trinh ticp TiTdng tU" X g . = - 2 X B , X c = - 2 x c . (1) tuyen cua (C) qua M. Gia su" ticp tuyen (t) kc lit M den (C) tiep xiic vdi (Cj tai N(x,; y,). Khi do phiTdng trinh cua (t) co dang Bay gicf ke't luan cua b^i toan sc dtfdc chufng minh nhd nhan xet sau: y - y , =(-3x| +6xi)(x-xi) Nhgn xet: X6t 3 diem A, B, C thuoc (C) co hoanh dp liTdng uTng la XA, XR, XC. Vi (t) di qua M ncn ta CO 6>: ' + Khi do A, B, C thang hang khi va chi khi XA + XB + Xc = 0. y o - y i =(-3xf +6x,)(Xo-x,). (1) ChuTng minh. Gi a suT A, B, C nam tren difdng thang co phU'dng trinh y = ax +b. Ngoai ra, do N thuoc (C) nen ta CO ^ Khi do XA, XB, XC la nghiem cua phU'dng trinh y, = - x ) ' + 3x^ - 2 • (2) x' - 3x + 2 = ax + b o x' - (3 + a)x + (2 - b) = 0 Nhir vay toa do tiep diem la nghiem cua he (1), (2). Ycu cau bai toan o he Ap dung djnh li Vi-ct, ta suy ra XA + XB + Xc = 0. (1), (2) v('Ji an la (x,; y,) co nghiem duy nhat.. NgiTdc lai, gia suT XA + XB + Xc = 0. Viet phifdng trinh difdng thang di qua A, B Thay y, tif phiTdng trinh (2) vao phi/dng trinh (1), ta di/dc c^t (C) tai C thi theo phan thuan ta co XA + XB + Xc = 0 suy ra Xc- = Xc suy ra 2x-| - 3(x,) + l)xf + 6x„x, - 1 -y,) = 0 C' trilng C va co nghla la A, B, C thang hang. Nhan xet diTdc chuTng minh. Quay trd lai bai toan, do A, B, C th^ng hang ncn theo nhan xet, ta co Lai thay 2xi'-3x„x?+xil-3xf+6x„x,-3xi^-0 ^ • XA + XB + Xc = 0. Theo (1), ta co XA- + XB- + X c = -2(XA + XB + XC ) = 0. o(x,-x„)2(2x,+x,)-3) = 0 (3) Tiep tuc ap dung nhan xet ta suy ra A', B', C th^ng hang (dpcm). Ta thay he (1), (2) co nghiem duy nhat khi va chi khi ~ ^ = x,, o x,, - 1 . Bai 3. Cho ham so' y - — - 3x - - CO d6 thj (C). Ttrdo tinh difdc y„ = 0. 2 X Vay M(l; 0) la diem duy nha't tren (C) ma qua 66 co the kc dung mpt Uep a) Khao sat va ve do thj (C). tuyen vdi (C). b) ChuTng minh rSng ham so co ba diem cifc tri phan biet A, B, C. , rir;. Bai 2. Cho ham so y = - 3x + 2. (C) c) Tinh dien ti'ch tarn giac ABC. a) Khao sat su" bien thien va ve do thj (C). d) Tim tam va ban kinh di/dng tron ngoai tiep tam giac ABC. b) Xet 3 diem A , B, C ihdng hang va thuoc (C). Gpi A', B', C' la giao diem cua TFa CO' y . = x - 30+ —1 = x-^-3x^+1 Htf(?ng d i n giai X (C) vdi tiep tuyen cua (C) lai A, B, C. ChiJng minh rang A', B', C th^ng hang a) ; c) Tim tren do thj (C) cac diem doi xuTng nhau qua 1(0; 2) Tird6y' = 0 o x ' - 3 x ^ + 1 = 0 . (1) • vi,;.V Hi^cJng d§n giai Dat f(x) = x' - 3x' + 1 thi f(-l) = - 3 , f(0) = 1, f(l) = - I , f(3) = 1 nen theo tinh a) Hocsinhtirkhiiosal. ^ ' chat ham lien tuc, phiTdng Irinh y' = 0 c6 3 nghiem XA, XB, XC thoa man dieu b) Phirdng trinh tiep tuyc'n cOa (C) tai diC'm A(XA; y^) c6 dang ki^n -1 < XA < 0 < XB < 1 < Xc < 3. Tir do suy ra dpcm. y = (xx-3)(x-XA) + yA.
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