Luyện thi Toán học - Cấp tốc giải 10 chuyên đề 10 điểm thi môn Toán: Phần 1

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Phần 1 tài liệu Cấp tốc giải 10 chuyên đề 10 điểm thi môn Toán giới thiệu tới người đọc các nội dung: Một số bài toán thường gặp về đồ thị, phương trình lượng giác, phương trình, hệ phương trình, bất phương trình đại số. Mời các bạn cùng tham khảo.

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Nội dung Text: Luyện thi Toán học - Cấp tốc giải 10 chuyên đề 10 điểm thi môn Toán: Phần 1

Cty TNHH MTV DVVH Khang Vi?t Cac em hoc sinh than men! Chuyen I Giang dirang Dai hoc la nci gan nhir em nao cung muon dirge birac vao sau M O T S O B A I T O A N T H U O N G G A P V E D 6 TH| khi rai THPT, dij biet rang khong phai tat ca deu c6 kha nang, dieu kien de thirc hien u-ac ma ay. Nham gop phan giiip cac em hoc sinh thirc hien du-gc ma irac 'i> Chii de 1: Tiep tuyen ciia d6 thi ha m so. v d) t do, tac gia bien soan cuon "Cap td'cgidi 10 chuyen de 10 diem thi mon Todn". Cuon sach du-gc bien soan theo noi dung chuan kien thirc, ky nang. Trong - Viet phuang trinh tiep tuyen khi biet toa do tiep diem M(xo;yo), hoac hoanh sach du-gc trinh bay tirng van de, theo cau triic de thi cua Bo giao due va Dao do X Q , hoac tung dp yg. ,f tao de ban doc tien tham khao. Moi van de se c6: - Viet phuang trinh tiep tuyen khi biet tiep tuyen di qua diem A{xf^;yf^) cho - Tom tat cac kien thirc ly thuyet ca ban. truac. - Lai giai chi tiet cac dang toan thirang gap va vi du minh hoa. I - Viet phuang trinh tiep tuyen khi biet he so goc cua no. - Cac bai tap ren luyen ky nang, c6 hirang dan chi tiet. • Phiro-ng phap: '' Tac gia chu tru-ang tranh du-a vao sach nhij-ng phan ly thuyet nang ne va it su- dung. Moi vi du, lai giai lai c6 nhan dinh sau sac, kem theo lai binh khien Cho h^m so y = f ( x ) c6 do thi (C) va M(xo;yo) la diem tren (C). Tiep tuyen nguai doc tam dac va se c6 tu- duy sang tao rieng cua minh khi gap nhCrng cau v a i d o t h i (C) tai M(xo;yo) c6: hoi kho, bai toan kho la khac. • Hesogoc: k = f'(xo) Phan bai tap t g luyen du-gc tac gia bien soan rat cong phu va tap hgp nhieu dang toan hay, mai me. Giiip nguai hoc khong chi c6 the thir sue nhirng bai • Phuangtrinh: y - y o = k ( x - X o ) , h a y y - y o = f ' ( x o ) ( x - X o ) toan ren luyen tir duy, ma eon giai mot each de dang nhCrng bai toan hoe biia, Vay, de viet duge phuang trinh tiep tuyen tai M(xQ;yg) chung ta can du ba yeu tuang chirng khong the nao giai noi. Mot so bai tap c6 the kho nhung each giai to sau: duge dga tren nen tang kien thuc va ky nang ca ban. Tac gia hi vong, khi gap • Hoanh do tiep diem: Xg mot de thi kho, la nguai hoe se khong eon ngai ngung trong viee dira ra lai giai cho moi bai toan. • Tung do tiep diem: yg (Neu de chua cho, ta phai tinh bang each thay Xg vao Cuon sach la sg ke thira nhirng hieu biet chuyen mon va kinh nghiem cua chinh hamso y o = f ( x o ) } • tac gia trong qua trinh true tiep dung lap boi duang.Tae gia hi vong, nguai hoc • Hesogoc k = f'(xo) can phai nam virng kien thue can ban truac khi tham gia bai tap t u luyen. De dat hieu qua cao trong ki thi Dai hoc sap tai, tac gia hi vong cac em su Dang 1. Viet phiro-ng trinh tiep tuyen Idii biet toa do tiep diem dung cuon "Cap td'cgidi 10 chuyen de 10 diem thi mon Todn" song song vai Viet phuang trinh tiep tuyen khi biet toa do tiep diem M(xo;yo), hoac hoanh cuon "Luyen thi cap toe mon toan" ma tac gia sap an hanh, phan bai mai eo nhieu bai toan mai la va eo nhieu each giai hay va dep. Cac em hoc sinh don do X Q , hoac tung do yg. doc. Hai cuon sach duge vi nhu moi ca the di tim mot nua kia trai tim, mong Bdi todn 1. Viet phuang trinh tiep tuyen cua do thj ham so y = f(x) tai diem muon ket tinh tinh yeu that dep. Moi cuon sach, mot phong each, mot net dep!. Vai "sang tac doc dao va day t u duy ngau hung" nay, tac gia tin tuang cac em hoc sinh se vi>ng tin buac vao ky thi Dai hoc sap tai. Sinh vien su pham va Gi&i. Tiep tuyen ciia do thi ham so y = f(x) tai M(xo;yo) 1^: 'i? rtir;~ .( ; , quy thay eo se eo them nhieu tai lieu de tham khao. y = f ' ( x o ) ( x - X o ) + yo SdaM.v...^ Mac dii tac gia da danh nhieu tam huyet cho cuon sach, song su sai sot la Bdi todn 2. Viet phuang trinh tiep tuyen ciia do thj ham so y = f (x) biet hoanh dieu kho tranh khoi. Chiing toi rat mong nhan duge su phan bien va gop y quy bau cua quy doc gia de nhirng Ian tai ban sau cuon sach duge hoan thien han. do tiep diem x = Xo. Tac gia Giai:Tinh yg =f(xo), y'(xo)=> phuang trinh tiep tuyen: y = f ( x o ) ( x - X o ) + yo
Cty TNHH MTV DVVH Khang Vi$t Ca'p te'c giai 10 chuyen 10 die'm thi m6n Toan - Nguygn Phu Khanh T a c o : y ' = 3 x ^ - 2 ( m - - l ) x + 3m + l Bai todn 3. V i e t p l i i r a n g t r i n i i t i e p t u y e n cua d o t h i h a m so y = f ( x ) biet t u n g Vai x = l = > y ( l ) = 3m + l = > y ' ( l ) = m + 6 dp tiep d i e m bang y o - Phu-ang t r i n h t i e p t u y e n t a i d i e m c6 x = 1 : y = ( m + 6 ) ( x - 1 ) + 3 m + 1 ' G i a i , Gpi M ( x o ; y o ) la t i e p d i e m Tiep tuyen nay di qua A ( 2 ; - l ) n e n c o : - l = m + 6 + 3m + l < » m = - 2 Giai p h u - a n g t r i n h f ( x ) = yg ta tirri du-p-c cac n g h i e m Xg. ,^ Vay, m = - 2 la gia t r i can t i m . ^ ' ^ ^ " ~" ^"" ' Tinh y'(xo) p h u - o n g t r i n h t i e p t u y e n : y = f ' ( x o ) ( x - X o ) + yo D a n g 2. V i e t p h i r c n g t r i n h t i e p t u y e n k h i b i e t t i e p t u y e n d i q u a V i d u 1. V i e t p h u ' a n g t r i n h t i e p t u y e n ciia d o t h j h a m so: y = x^ - 3 x ^ + 2 d i e m c h o tru'o-c 1. T a i d i e m ( 2 ; - 2 ) 2. T a i d i e m c6 h o a n h d p x = - 1 3. T a i d i e m c6 t u n g d p y = - 2 4 . T a i g i a o d i e m ciia d o t h i v o i y = x - 1 V i e t p h u ' a n g t r i n h t i e p t u y e n k h i b i e t t i e p t u y e n d i qua d i e m h{y-p,;yp,) chotru-ac Lffi g i a i G i a i : Gpi M ( x o ; y o ) la t p a d p t i e p d i e m . K h i d o t i e p t u y e n c6 d a n g : if,;* , i H a m so d a c h o xac d j n h vo-i V x e M . y = f ' ( x o ) ( x - x o ) + yo Gpi M o ( x o ; y o ) la t p a d p t i e p d i e m v a y g = y ( x o ) = Xo - S x g + 2 V i t i e p t u y e n d i q u a A n e n c6: y ^ = f ' ( X Q ) ( X A - XQ ) + y g , g i a i p h u ' a n g t r i n h n a y y ' = 3 x ^ - 6 x , tiep tuyen tai diem M Q c o h e s o g o c : y ' ( x o ) = 3 x o - 6 X Q ta t i m d u ' a c Xg, s u y ra p h u ' a n g t r i n h t i e p t u y e n . 1. T a c o : X Q = 2 = > y ' ( 2 ) = 0 . x + 2 Phu'ang t r i n h tiep tuyen tai d i e m M ( 2 ; - 2 ) : y = 0 ( x - 2 ) - 2 = - 2 V i d u 1. V i e t p h u - a n g t r i n h t i e p t u y e n d v a i d o t h j ( C ) : y = , biet d d i qua X ~~ 2 2. Ta c6: X Q - - 1 => yo = - 2 , y ' ( - l ) = 9 diem A(-6;5). Phu'ang t r i n h t i e p t u y e n : y = 9 ( x + l ) - 2 = 9x + 7 L a i gi^i. 3. T a c o : y g = 6 = > x ^ - 3 x ^ + 2 = - 2 < = i > x ^ - 3 x ^ + 4 = 0 C a c h 1: Gpi ( x g ; y ( x o ) ) la t p a d p t i e p d i e m c i i a t i e p t u y e n d v a ( C ) , v a i ,. ( X Q + l ) ( x o - 2)^ = 0 XQ = - 1 h o a c Xg = 2 y ( x g ) = ^° """^ , t i e p t u y e n d c6 h e so goc y'(xo) = ~~2' Xg 5^2 v a d c6 >^o-2 (xo-2) Phu-ang t r i n h t i e p t u y e n t a i d i e m ( - l ; - 2 ) : y = 9x + 7 Phu'ang t r i n h tiep tuyen tai diem ( 2 ; - 2 ) : y = - 2 phu'ang trinh: y = ^ — Y ( x - X g ) + ^° (xo-2) ^0-2 Vay, CO 2 t i e p t u y e n t h o a m a n de b a i : y = - 2 , y = 9 x + 7 . 4. P h u a n g t r i n h h o a n h d p g i a o d i e m : x^ - 3 x ^ + 2 = x - 1 d di qua diem A(-6;5) n e n c6 5 = ^ — - ( - 6 - X o ) + ^° phu'ang trinh (xo-2f X o - 2 o x ^ - 3 x ^ - x + 3 = 0 < : i . ( x - 3 ) ( x ^ - l ) = 0x = 3 h o a c x = ± l n a y t u - a n g d u ' a n g v a i Xg - 6xg = 0 Xg = 0 h o a c Xg = 6 Phu'ang t r i n h tiep tuyen tai d i e m ( - l ; - 2 ) : y = 9x + 7 * V a i Xg = 0 , ta CO p h u - a n g t r i n h : y = - x - l s fii|;::•.' Phu'ang t r i n h tiep t u y e n tai d i e m ( l ; 0 ) : y = - 3 x + 3 x 7 'ik^yxi'-v' * V a i Xg = 6 , t a CO p h u - a n g t r i n h : y = — + - Phu'ang t r i n h tiep tuyen tai diem ( 3 ; 2 ) : y = 9 x - 2 5 .v 4 2 ficji.. V a y , CO 3 t i e p t u y e n t h o a m a n d e b a i : y = 9 x + 7, y = - 3 x + 3, y = 9 x - 2 5 . Vay, CO 2 t i e p t u y e n t h o a d e b a i y = - x - 1 , y = - - + - . 4 2 ' i d u 2. C h o h a m so: y = x ^ - ( m - l ) x ^ + ( 3 m + l ) x + m - 2 . Tim m de tiep C a c h 2: P h u - a n g t r i n h d d i q u a A ( - 6 ; 5 ) c6 h e so goc k , k h i d o d c6 p h u - a n g i y e n cua d o t h j h a m s o t a i d i e m c6 h o a n h d p b a n g 1 d i q u a d i e m A(2;-l). trinh l a : y = k ( x + 6) + 5 Lo-i g i ^ i d tiep xuc ( C ) t a i d i e m c6 h o a n h d p Xg k h i va c h i k h i h e : - ji'^-^ H a m s o da c h o xac d i n h v a i V x e R . • 5
Ca'p ta'c giai 10 chuyen dg 10 diem thi m6n Toan - NguySn Phii Khanh Cty TNHH MTV DVVH Khang Vi$t 2. Tim gia tri m e R de tiep tuyen tqi M j , M2 song song v&i nhau. k(xo+6) + 5= " 4xg-24xo=0 Xg -2 2x + 1 CO nghiem XQ hay 4 CO nghiem Xg k = - V i d u 2 . Viet phu-ang trinh tiep tuyen d ciia do thj ( C ) : y = biet d each k = - X H" 1 deu 2 diem A(2;4) va B ( - 4 ; - 2 ) . Xo=0, k =-l=>d:y = - x - l Lo-i gi^i . • d - ^Xt' m Goi M ( x o ; y ( x o ) ) , XQ 5 ^ - 1 la toa do tiep diem ciia d va (C) Xo=6, k =- - = > d : y =- - + - X 7 Khi do d CO he so goc y ' ( x o ) = j va c6 phu-ang t r i n h la : ^ Vay, CO 2 t i e p t u y e n t h o a d e b a i y = - x - l , y = — + - . (xo+1) 4 2 * Nhan xet 1: Qua each 1 ta thay du-ong thang d : y = - x - 1 luon tiep xiic vo-i (C) tai tiep diem M ( 0 ; - l ) va du-o-ng t h i n g d luon vuong goc v a i du-ang thang Vi d each deu A, B nen d d i qua trung diem cua AB hoac cung IM v a i I l a g i a o d i e m 2 du-ang tiem can. phu-ang v a i A B . •'^'.'.•r ,//^ >-'•>?•;;-;.;,•,:.•• Qua do ta c6 bai toan saw. X + 2 T H I : d di qua t r u n g diem l ( - l ; l ) , t h i ta luon c6: W A * u p ,viV/ Tim treii do thi y = nliimg diem M sao cho tiep tuyen tqi M vuong goc v&i x-2 1= — - XQ ) + 2 — , phu-ang t r i n h nay c6 nghiem Xg = 1 = y (Xg+1) Xg+1 iC.„ , ... ;v .. - " X i ; . dwang thang I M , v&i l ( 2 ; l ) . Gg-iy. Vai xg = 1 ta c6 phu-ang t r i n h tiep tuyen d : y = - x + - . H >Md 9
ca'p tfiic g i i i 1 0 chuySn d l 1 0 difi'm thi m 6 n Toan - Nguygn Phu Khanh Cty TNHH M T V D V V H Khang V i ? t m = -4 1. Tiep tuyen c6 he so goc bang - 1 . m = 4 2. Tiep tuyen song song v a i d u a n g t h i n g d : y = - 4 x + 1 . 3. Tiep tuyen tao v a i 2 true tpa dp lap thanh mot tam giac can. • ' - ' hay < 20V3 m = -- 4. Tiep tuyen tai diem thupc do thj c6 khoang each den true Oy bang 2 . L a i giai EL;); m = 4xo - 8 x^0 2073 r4l m =- Ham so da cho xac dinh v a i Vx 1. Ta c6: y ' = ^—j 3..B'..'L^?. Vo-i m = - 4 , tiep tuyen y = - 4 x + 4, tiep diem lVIj(l;0) KWgitfJ (^-l) Vo-i m = 4 , tiep tuyen y = 4x + 4 , tiep diem M2 (-1;0) Go! M ( x o ; y o ) la tpa dp tiep diem, suy ra phu'ang t r i n h tiep tuyen cila (C): -4 , s 2xo+2 - -4 . 2xo+2 20^3 ^ . 2 0 ^ . . . . ..f 16 y =- _ ( x - X o ) + ^ ^ v a i y (xo) = - -j va yo Vm m = , tiep tuyen y = x + 4 , tiep diem Ivl 3 ' 9 1. Tiep tuyen C O he so goc b a n g - 1 , ,> . , H , .. > 20N/3 ^ . 20S . ^ . v ... N/3 16 Vai m = — ^ — , tiep tuyen y = — - — x + 4 , tiep diem M -4 • > 3 '9 Nen c6: = - 1 XQ = 3, X g = - 1 h q?j,U 9h &'(>. Vay, qua A ke du-gc 4 tiep tuyen den do thj ( C ) : * V a i XQ = - 1 ^ yo = 0 A:y - - x - 1 | K B - I - fst ~ [ ^ ^ . . . . 2073 , 2073 , , * V a i XQ =2=:J.yo = 4 = > A : y = - x + 7 ' '"^ Sccj y = - 4 x + 4 , y = 4x + 4 , y = —x + 4,y = x + 4 I- 9 9 Vay, C O 2 tiep tuyen thoa man de bai: y = - x - 1 , y = - x + 7. • Mff rong: Don^ todn qua 1 cf/em /ce cfu-p-c 4 t/ep twye>i t/en (fo thi la dang loan 2. Tiep tuyen song song v a i du'ang thang d : y = - 4 x + 1 . ' = f,^ j it gap. Dehieu kihan dang todn nay, tagidi bdi todn sou: "Bien luan theo m so tiep tuyen cua (C): y = x'*-6x^ v e t i r d i e m M(3;m)". Nen c6: y ' ( x o ) = -4 - - - 4 < : : > X o = 0 hoac XQ = 2 ^ ^ ! • Gg-iy: Phu-cng trinh tiep tuyen ( d ) cua (C) ve tij- M(3;m) c6 dang: y=k(x-3)+m * Vai XQ = 0 => yo = 2 = ^ A : y = - 4 x + 2 Wt> intra r': (d) va (C) tiep xuc nhau tai diem cop hoanh do X g , tCi' do suy ra: * Vai xo = 2 = > y o = 6 ^ A : y = - 4 x + 14 ..fy;.,,,,.^^ m = x ^ - 6 x 2 - ( 4 x ^ - 1 2 x o ) ( x o - 3 ) = g(xo) , 1 - roV Vay, C O 2 tiep tuyen thoa man de bai: y = - 4 x + 2, y = - 4 x + 1 4 . Ta c6: g'(xo) = 0 < ^ X Q = - 1 h o a c x o = l hoacxo=3 ^.uu.^ i j j aJO-o'< '^'^^''^ 3. Tiep tuyen tao v a i 2 true tpa dp lap thanh mot tam giac can nen he so goc cua tiep tuyen bang t l . M a t k h a c : y ' ( x o ) < 0 , nen c6: y ' ( x o ) = - l , Tir bang bien thien suy ra: ^ ^ |b «6fn fiOiH nayuJ qSiJ ,y.i;V _4 i'>^'' * m < - 2 1 hoac m = 2 7 : C O 2 tiep tuyen. ;•; Tu-c j = -\oy.Q=-\\\' Tiep tuyen song song v a i du-ang thang ( d ) : ax + by + c = 0 4. Khoang each t u M(xo;yo) den true Oy bang 2 suy ra XQ = ±2 , suy ra M Tiep tuyen vuong goc v a i du-ang thang ( d ) : ax + by + c = 0 ' "'JJ b M(2;6). ..,,-'.,.:.r:-,r'^..:..v Tiep tuyen cung v a i du'ang thang (d):ax + by + c = Otao thanh goc cp. 2^ 4 2 2x + 2 Phu'ang t r i n h tiep tuyen tai M - 2 ; - la: y = — x - - ,,1 ; V i d u 1. Cho ham so: y = CO do t h i (C). 3 x-1 Phu-ang t r i n h tiep tuyen tai M(2;6) la: y = 4x + 14 ; r!?! V Viet phu'ang t r i n h tiep tuyen cua do thj ( C ) .
Ca'p ta'c giai 10 chuy§n de 10 dig'm thi mCn Toan - NguySn Phil Khanh Cty TNHH MTV DVVH Khang Vigt 4 2 Vay, CO 2 tiep tuyen thoa de bai: y = - - ^ - - - y = 4x + 1 4 . 3. (dO i - ( d 2 ) « k,k2=-1 ^ ^^^^^^ =p COSP = JM^l Vi du 2. T i m tat ca cac gia t r j ciia k de ton tai 2 tiep tuyen v o i (C): Vkf+l.^k hi • Gia Sit A ( x i ; y i ) , B(x2;y2)/ ditang thang di qua hai diem phan biet A, B c6 he y = + 6 x ^ + 9 x + 3 phan biet va c6 cung he so goc k , dong tho-i du-ong thang di qua cac tiep diem cua 2 tiep tuyen do vo-i (C) cat cac true Ox, Oy s6g6ck = ^^^^, X2^xi. ' tu-ong li-ng tai A, B sao cho OA = 2012.OB. B a i t a p tu- l u y e n : , , >v:.„,, Lo-i giai 2x Hoanh do tiep diem Xg cua tiep tuyen dang y = kx + m v a i (C) la nghiem cua \t phu'ang t r i n h tiep tuyen v a i do t h i ham so: y = , biet: »•:* ; x-1 phu-ongtrinh f ' ( X Q ) = k « Sxg + 1 2 x o + 9 - k = 0 ( l ) a. He so goc ciia tiep tuyen bang - 2 .^^j^ b. Tiep tuyen song song voidu-ong thang ( d ) : x + 2y = 0 , , De ton tai 2 tiep tuyen v a i (C) phan biet nhau t h i p h u a n g t r i n h ( l ) c6 hai c. Tiep tuyen vuong goc v a i du-ang thang ( A ) : 9 x - 2 y + 1 = 0 ; • nghiem phan biet, khi do A' = 9 + 3 k > 0 hay k > - 3 ( 2 ) . Hiro-ng dan giai Khi do, tpa dp tiep diem ( x g i y o ) cua 2 tiep tuyen v o i (C) la nghiem he 2(x-l)-2x -2 ' Ta c6: y ' = -^^ ' - ^ = \> yo = 5 ^ 0 + 6 x 0 + 9 x 0 + 3 yo =^(xo +2)(3xg + 12xo + 9 ) - 2 x o - 3 phuong trinh: 3xo + 12xo+9 = k 3xo + 12xo+9 = k Gpi ( X Q ;yo) la tpa dp tiep diem, ta c6 he so goc tiep tuyen tai (xo ;yo) bang 1/ „x, „ , k-6 2k-9 yo = 3 ( X Q + 2)k - 2 X O - 3 = — x o + ^ — (^0-1) 3xo + 1 2 x o + 9 = k -2 a. Theo giai thiet, ta c6: y '(xn) = - 2 = -2 Vay phu-o-ng t r i n h du'mig thang di qua cac tiep diem la ( d ) : y = - y - x + • (xo-1)^ '..^'-i - Xo-l = l Xo-2=>yo=4 ,^(X!!r>' Do ( d ) cat true Ox,Oy tu-ang li-ng tai A va B sao cho OA = 2012.OB nen c6 the «(xo-lf = 1 0 xayra: k - X o - l =- l [xo=0^yo=0 Vay, CO 2 tiep tuyen thoa de bai: y = - 2 x + 8,y = - 2 x < - Neu A = 0 t h i B = 0 , tru-ong hop n ^ y c h i thoa neu (d) cung qua 0 . -2 1 7 1 Sf'i' ^ ^ = ±2012 =>k = 6042 OA 3 c. Theo giai thiet, ta c6: ~^ ^ = - ^ c » ( x o - l ) ^ =-^ S;?-SH'V hoac k = - 6 0 3 0 (khong thoa ( 2 ) ) . iVj 9 2 32 2 8 ! Vay k = - , k = 6042 thoa bai toan. Vay, CO 2 tiep tuyen thoa de bai: y = - - x + — , y = - - x + - - ill- Chuy: • : 2. Cho ham so: y = x^ - 3mx^ - x + 3m c6 do thi (C^,). Djnh m de (Cn,) tiep xuc • Cho hai duang thang {A'^)\y = k-^x + hva [A2)-.y = \i2^ + m. v a i true hoanh. , ,. , 1. { d i ) | | ( d 2 ) < = > j * ^ i "'^2 2. {A^),{d2) ciingphirang 'k^=k2 Hiro-ng dan giai (Cm) tiep xuc vai true hoanh tai diem c6 hoanh do X Q khi va chi khi he phu'ang trinh: 10
Ca'p te'c giai 10 cliuyen de 10 die'm thi mon Toan - NguySn Phu Khanh Cty TNHH MTV DVVH Khang Vigt X Q - 3 m x o - x + 3m = 0 , Xo(xo-l)-3m(x^-l) = 0 //I ^ _ 32 _64 32 64- CO nghiem X g tiix he * Vai x = - = > y ' v3y phu-ang t r i n h tiep tuyen y = x+ — 3xo - 6mxo - 1 = 0 3xo -6mxn0 •- 1 = 0 27 27 2x-l 5, Cho ham so y = , c6 do t h i (C). Viet phu-ang t r i n h tiep tuyen cua (C). (xo'-l)(xo-3m) = 0 , 1 x-1 CO nghiem X g ^ ' CO nghiem X Q => m = ± - 3x^-6mxo-l = 0 ^ a. Tiep tuyen CO he so goe b a n g - - . , v , 3. C h o h a m s o : y = x'^+x^+(m-l)x^ - x - m c6 do t h i ( C „ T ) . D i n h m de (C„) b. Tiep tuyen tao v a i hai tiem can mot tarn giac CO chu vi nho nhat. ' tiep xuc v a i true hoanh. - : c. Khoang each t i l - l ( l ; 2 ) den tiep tuyen Ian nhat. , j £ ^ Jvf., 'N- H i r a n g dan giai d. Tiep tuyen ciia (C)tai IVI v u o n g g o c v a i IM. (Cn,) tiep xuc v a i true hoanh tai diem c6 hoanh do X Q k h i va chi k h i he phu-ang Hu-6-ng dan g i a i . I-),-•'••;'!• s"! ..• ,; x^ + x ^ + ( m - l ) x ^ - X o - m = 0 ^ Gpi M ( x g ; y o ) la toa dp tiep diem cua (C) va tiep tuyen ( d ) , , ,,,,,;,|^, j. trinh: CO nghiem Xg 4xo+3xo+(m-l)xo-l = 0 Tiep tuyen (d) tai M ( x g ; y o ) : y = - 1 _5 -r (xx- -xX g ) u+ 2 ^ 0 - 1 i ( x ^ l ) + Xo Xg-l+ m(xo-l) = 0 ^ ' CO nghiem X g a. — ^ - " ' " ^ ' ^ X g -3,Xg - - 1 4xo + 3xo + ( m - l ) x o - 1 = 0 (xg - i j ^ X g + X n + mi )| = 0 1 Tu-do ta t i m du-ac tiep tuyen la: y = - i x + — v a y = - — X + — . ' CO nghiem Xg => m = - 2 , m = 0, m = - 4 4 4 4 4xo + 3xo + ( m - l ) x o - 1 = 0 2x 0 b. Tiep tuyen (d) cat tiem can du-ng tai A , cat du-ang tiem can ngang 4. Cho ham so y = (2 - x)^ x^, c6 do thi (C). Viet phu'ang t r i n h tiep tuyen cua (C): a. Tai giao diem ciia (C) vai Parabol y = x^. tai B ( 2 x o - l ; 2 ) ^ I A = - ^ , I B = 2 | x g - l | ^ l A . I B = : 4 . ^ Xg - 1 b. Tiep tuyen di qua diem A ( 2 ; 0 ) . Chu vi tarn giac lAB : p = AB + lA + IB = V l A ^ + I B ^ + lA + IB H i r a n g dan giai , , y = x'^-4x^+4x^=>y' = 4x^-12x^+8x \i in oi';{*'•> ^ Mat khac: lA^ + IB^ > 2IA.IB = 8, lA + I B > 2VIA.IB = 4 . Nen p > 2>/2 + 4 a. Phu-ang t r i n h hoanh do giao diem: Dang thu-c xay ra khi: lA = IB ( x g - 1 ) ^ = 4 Xg = S.Xg = - 1 x ' ^ - 4 x ^ + 4 x ^ = x 2 « x ^ ( x ^ - 4 x + 3) = 0 o x = 0,x = l , x = 3 Tir do ta t i m du-ac tiep tuyen la: y = - i x + ^ va y = - i x + -^ \ * Vai x = 0 => phu-ang t r i n h tiep tuyen: y = 0 c. Goi H l a h i n h c h i e u c u a l i e n ( d ) , t a c 6 : d ( l , A ) = IH j .4 9. * Vai X = 1 => phu-ang t r i n h tiep tuyen: y = 1 * Vai X = 3 => phu-ang t r i n h tiep tuyen: y = 2 4 x - 6 3 AUi Trong tarn giac vuong lAB, c6 : - A r = ^ +-^r> = - . Suy ra: IH < V2 . IH2 IA^ IB^ IA.IB 2 ^ . b. M ( x g ; y g ) 6 ( C ) . T i e p tuyen ( t ) cua (C) tai M ( x g ; y o ) : Dang thii-c xay ra c:> lA = IB ^' ''' y = (4x^-12x^+8xg)(x-Xg) + x^(xo-2f irm'SirdM'nImM'< Tir do ta t i m du-ac tiep tuyen la: y = -—x + — va v = - — x + — 4 4 ^ 4 4 A ( 2 ; 0 ) e ( t ) o ( 2 - x g ) ( 3 x ^ - l O x ^ + Sxg) = 0 Xg = 0 , X o = 2 , X o = ^ i'>^ -1 Du-ang thang (d) c6 vecta chi phu-ang u = 1;- * Vai xg = 0 = > y ' ( 0 ) = 0,yg = 0 = > phu-ang t r i n h tiep tuyen y = 0 * Vai X g = 2 = > y ' ( 2 ) = 0,yg = 0 = > phu-ang t r i n h tiep tuyen y = 0 Vj IM = , IMl(d)xo-l- - = OCi> X g = 0 , X g = 2 Xn -1 12 /
C&p tO'c g i i i 10 chuyfin d6 10 di6'm thi mOn Toan - NguySn Phu KhAnh Cty TNHH M T V D W H Khang Vi$t Tu- do ta t i m dirge tiep tuyen; y = - x + l , y = - x + 5 (d) la tiep tuyen ehung eua (C) va ( C ) nen d [ l , d ] = R, tii- day ta t i m du-p-c 6. Viet phu'cng trinli tiep tuyen (d) ciia (C) tai diem M thupc (C): phu-ang t r i n h : ( X Q + 1 ) ' ' ^ - 5 ( x o + 1)^+4 = 0 , 17^ 2x-3 y = x'* + 2x^ - 1 sao cho ( d ) vuong goc v a i A M , biet A 0;- 9. Gpi M i a diem bat ki tren (C) eua ham so y = va I la giao diem 2 V 8, x-2 H i r a n g d a n giai du-ang tiem can. Tiep tuyen (d) cua (C) tai M cat 2 du-ang tiem can tai A va Goi M ( x o ; y o ) la toa do tiep diem cua (C) B. Tim tpa dp diem M sao cho du-ang t r o n ngoai tiep tam giac lAB c6 dien tich bang 271. 25^ AM = Xo;xo +2XQ - , du-ang thang (d) tiep tuyen tai iVl c6 vecta phap Hiro-ng dan giai Kfrfni^i 8 M J l Eli 2xo-3 tuyen la n = (4x^ + 4 x o ; - l ) . Theo bai toan, ta c6 A M va n cung phu-o-ng, tir Goi M XQ;- ,Xo 7^2 la diem thupc (C). Xn -2 daytimdu-o-c M ( 0 ; - 1 ) , M(-l;2), M(1;2). •7 Tiep tuyen ( d ) tai M c6 he so goc la: y ' ( x o ) = 7. Viet phu-ang t r i n h tiep tuyen d ciia (C): y = - i x ' ^ - x ^ + | , biet d cat true (xo-2f -1 2x —3 hoanh , true tung Ian lu-gt tai A va B sao cho O A > 0 B , sao cho dien ti'ch tam Phu-ang t r i n h eo dang: ( d ) : y = ^ ( x - X o ) + —2 , ( d ) cat 2 tiem can (xo-2) Xo-2 giae G A B bang ^ va khoang each tir 0 den d bang ^ 2xo-2 Ian lu-at tai: A 2; , B(2xo-2;2) Hiro-ng dan gi^i iq"'»- Xn-2 Gpi H la hinh chieu vuong goeeua 0 len d t h i OH = d ( 0 ; d ) XQ = l = > M ( l ; l ) AIAB vuong tai I va S,AB = 27t (XQ - if + —- =2 i0A.0B = - XQ = 3 ^ M ( 3 ; 3 ) 2 2 0A.0B = 3 OA = 3 Ta c6: 1 1 10 OA^+08^=10 [OB-l 10. Lap phu-ang t r i n h tiep tuyen ciia do t h i (C) : y = ^ ^ ^ tai nhu-ng diem thupc -+ OA^ OB^ OH^ 9 OB do thj CO khoang each den du-6-ng thang ( d ) : 3x + 4 y - 2 = 0 bang 2. He so goc ciia du'ang thang d la k = ± — = ±3 OA Hu-o-ng dan g\&\ Tiep tuyen d : y = - 3 x +3, y = 3x + 3 . Gpi M ( x o ; y o ) la diem thupc do t h i ( C ) , k h i d 6 : =y(xo) = ^ ^ ^ ^ x+3 8. Viet phu-o-ng t r i n h tiep tuyen d ciia (C): y = tai diem M sao cho tiep Taco: d [ M , ( d ) ] = 2 3xo+4 -12 = 0 o 3x5-Xo=0 Xo=0 4 V 5^0 + 1 Du-ang t r o n ( C ) c o t a m l ( - l ; l ) , ban kinh hoac X g = - 2 Tiep tuyen ( d ) tai M 1 ; - thupc (C) CO phu-ang t r i n h : 2xo + 3 l (xo + l ) T H 2 : 3xo + 4yo + 8 - 0 3xo + 4 + 8 = 0 •»3x^+19x0+20 = 0 x , -+1 2 - ^ ( x - x o ) + y - l - ^ =0 (xo+1) '^o + l
ca'p tOc giai 10 chuySn ai 10 djgm thi m6n Toan - Nguygn Phu Khanh Cty TNHH MTV D W H Khang Vi$t 5 , . 1 5 Phiro-ng t r i n h tiep tuyen (d) tai M tliuoc do t h i (C) c6 dang: m < — hoac — < m < — 2 6 18 * y = y ' ( x o ) ( x - x o ) + y ( x o ) trong do va y ' ( x o ) = 12. T i m tham so m de parapol ( P ) : y = x ^ - x + l va do thj (Cn,) cua ham so (xo+1)- mil} rA W f!!>3:i* y = x'^ - 4mx^ + 7mx - 3m tiep xuc nhau. Phu'ang t r i n h tiep tuyen ( d ^ ) tai (0;3) la y = - x + 3. Hiro-ng d a n g\M 1 11 47 (P) va (C^) tiep xuc nhau tai diem c6 hoanh do Xg khi va chi khi he phu-o-ng Phu-ang t r i n h tiep tuyen (d2) tai M2 la y = x+ s'4 16 16 X Q - X Q + 1 = Xg - 4mxo + 7mxo - 3m 7 1 23 trinh: CO nghiem Xg Phu'cng t r i n h tiep tuyen ( d j ) tai M 3 -Si- la y = x+ — . 2xn - 1 = 3xn - 8mxQ + 7m 4 16 16 (xg - l ) ( x o - 4mxo + 1 + 3 m ] = 0 Phu-ang t r i n h tiep tuyen ( d 4 ) tai M4 la y = - 9 x - 1 3 i ' CO nghiem Xg 3 2xo - l = 3xo - S m x g + 7 m ( * ) Vay, CO 4 tiep tuyen thoa de bai: 9 47 1 23 _ * Vai X Q = 1 thay vao phu-ong trinh ( * ) , ta du-o-c m = 2 ^-h y = - x . 3 , y = - - x . - , y = - - x . - , y = -9x-13. * Voi Xg - 4 m x o + 1 + 3m = 0 ket ho-p phuong trinh (*) ta du-gc he: 11. T i m tat ca cac diem tren du-ong thang 30x - 24y + 61 0 de tij- do ke den do t h i m +1 1 Xn = ,m ^ - „3 2 y 2xo-l = 3xo-8mxo+7m ° 2m-l 2 (C) : y = + 2x + - 3 tiep tuyen tu-ong u-ng v a i 3 tiep diem c6 hoanh 2 3 2 3 ^ X Q - 4mxo + 1 + 3m j.3 = 0.3 = 0 ^ f m + 1 ^ f m+ 1 ^ do X]_,X2,X3 thoa X;^ < X 2 < 0 < X 3 . -4m + l + 3m = 0 Um-lv Um-lJ , . f < H i r a n g d a n giai 1 •(4m + l ) ( m - 2 ) ( m - l ) - 0 hay mG0 m < — hay m>— 2 6 the can phai chia ra tCrng t r u a n g hp-p de giai. Tac gia hu-ang dan 1 tru-ang hp-p, 3 12 5 tru-ang hp-p con lai gianh cho ban dpc. —-m>0 c*' m < — 18 18 ,x^-2x0-15 5 Gia su- ( d ) eat tiem can du-ng tai A eat tiem can ngang tai ^m-^
cap tO'c giSi 10 chuyfin 6& 10 die'm thi mOn Toan - N g u y j n Phu Khanh Cty T N H H M T V DVVH Khang Vi?t lA De do t h i cat true hoanh tai 4 diem phan biet M, N, P, Q k h i va chi k h i a. Tiep tuyen (d) c6 he so goc la k = ± — = +4, suy ra c6 2 tiep tuyen. IB A' = 4m^ - 4 m > 0 o 2 phu-ang t r i n h (2) c6 2 nghiem phan biet du-o-ng P = 4 m > 0 hay b. IA = , IB = | x o + 3 | , t h e o b a i t o a n t a s u y r a ( x o + 3 ) =4 Xo+3 S = 4m>0 m>l. "i> C h i i d e 2: T u o n g g i a o g i u a h a i d6 thi. Vo-i m > l phu-ang t r i n h (2) c6 2 nghiem phan biet t^ = 2 m - 2 V m ^ - m hoac • Lap phu-o-ng t r i n h hoanh do giao diem cua hai do t h i (C): y-f(x) t2 = 2m + 2vm^-Hm , gia s u - t j < t 2 . (C'):y = g ( x ) l a : f ( x ) = g ( x ) ( * ) . • Bien luan so nghiem cua phu-ong t r i n h (*), so nghiem phu-o'ng t r i n h {*) la so jVIQ = 2NPOQ^ = 40P^ tii-c la t2 = 4 t i sVm^-m = 3 m , binh p h u a n g 2 ve giao diem cua (C) va (C). va r u t gpn ta du-gc phu-ang t r i n h 16m^ - 2 5 m = 0, phu-ang t r i n h nay eo m = — 16 V i du 1. thoa dieu kien m > 1 . 1. T i m m de du-o-ng t h i n g d : y = - x + l cat do thj (C) ham so Vi du 2. y = 4 x ^ - 6 m x ^ + l tai 3 diem A ( 0 ; l ) , B, C saocho: 1. Cho ham so y = c6 do t h i la ) . T i m m de t r e n do thj (C^) eo 2 a. B, C doixu-ngqua y = x b. 0B.0C = - 4 X 1 2. T i m m d e h a m s o y = - x ' * + 4 m x ^ - 4 m cat true Ox tai 4 diem phan biet M, diem P, Q each deu 2 diem A ( - 3 ; 4 ) , B ( 3 ; - 2 ) va dien tich tii-giac APBQ N, P, Q (XM PQ = - j 2 ( x 2 - X J ) ^ 2. Phu-o-ng t r i n h hoanh do giao diem cua ham so va true hoanh: Dien tich tu-giac APBQ bang 24 -x'^+4mx^-4m = 0 ( l ) . Dat t = x^, (t>0) k h i do phu-o-ng t r i n h ( l ) tra d(A;PQ).PQ = 24 O 3 A ^ ^ 2 ( X 2 - x ^ f = 2 4 c ^ ( x i + X 2 f - 4 x i X 2 = 16 (2) thanh - t ^ + 4 m t - 4 m = 0 , t > 0 (2). Theo djnh ly Vi - e t , ta eo: x^ + X2 = m, X j .X2 = - 3 19
Cap te'c giai 10 chuyen 10 di§'m thi mOn Toan - MQuyfin I'iiu Khanh Cty TNHH WITV DVVH Khang Vig Thay vao (2) ta du-o-c +12 - 1 6 = 0 m = - 2 hoac m = 2 Lai c6: MN = ^j{x2 - X j ) ^ + (x2 + m - X i - m ) ^ = ^(x2 + X i ) ^ - 4 x 2 . X i Doi chieu dieu kien, ta thay m = 2 thoa man bai toan. 2. Du'cmgthang ( d ) cophu'o-ngtrinh: y = m ( x - 2 ) . _ . . , Hay MN = ^ 2 ( m ^ - 2 m + 13) Phu-ang t r i n h hoanh dp giao diem ciia ( d ) va ( C ) : m-1 (d): x - y + m = 0 =>d[l,(d)' = - x ^ + 6 x ^ - 9 x + 2 = m ( x - 2 ) c : > ( x - 2 ) ( x ^ - 4 x + l + m) = 0 . Vo-i m > 0 , ( d ) cat (C) tai 3 diem phan biet A,B,C va A la trung diem EC khi Khi do: (**) c : > ^ 2 ( m ' ^ - 2 m + 1 3 J — - — = 8, binh phu-ang 2 ve roi r u t gon t va chi khi phu'ang t r i n h x ^ - 4 x + l + m = 0 (*) c6 2 nghiem phan biet X i , X 2 du'gc: ( m - l ) V l 2 ( m - l f - 6 4 = 0,dat t = ( m - l ) ^ > 0 , m>0 khac 2 dong t h a i m > 0 o ] A = 3 - m > 0 o 0 < m < 3 ta du-ac t^ + 1 2 t - 6 4 = 0 => t = 4 tux ( m - i f = 4 « • m = - 1 hoac m = 3. [m;t3 * Vai m = - 1 t h i phu'ang t r i n h (*) t r a thanh: x ^ - 4 x = 0x = 0 hoac x = 4 ; Vo-i 0 < m < 3 t h i (d) cat (C) tai 3 diem phan biet A(2;0), => M ( 0 ; - 1 ) , N ( 4 ; 3 ) hoac ngu'ac lai la toa do can t i m . B(2-V3-m;-mN/3-m), c(2 + V 3 - m ; m N / 3 - m ' . * Vai m = 3 t h i phu-ang t r i n h (*) t r a thanh: x ^ - 4 = 0x = - 2 hoac x = 2 BB'+ CC :=> M ( - 2 ; 1 ) , N ( 2 ; 5 ) hoac ngu-gc lai la toa do can t i m . Theo bai toan, Sggcc = B'C' = 8 my7 3 - m = 2 , binh phu'cmg 2 ve 2 Vay, M ( 0 ; - l ) , N ( 4 ; 3 ) hoac M ( - 2 ; 1 ) , N ( 2 ; 5 ) hoac ngu-gc lai la toa do can t i m . ,3 o „ 2 r o ii r u t gon ta du-ac m - 3 m + 4 = 0(m + l ) ( m - 2 ) = 0 < » m = - l hoac I. Du-ang thang ( d ) qua 0 c6 dang: y = kx, y = mx ( k , m deu khac 0 va k m ]. m = 2. Doi chieu dieu kien 0 < m < 3 , suy ra m = 2 thoa man de bai. Goi A,B la giao diem cua ( d ) va (C) thi toa do A,B la nghiem cua he: Vi d u 3. 2 2x + l y=- 1. T i m tga do M,N de du-ang thang ( d ) : y = x + m cat do thj y = tai 2 k. x-1 X ^ y = kx [y = kx diem phan biet M,N sao cho = 4 [ I la giao diem 2 tiem c a n ) . 2 De A,B ton tai t h i k > 0 , khi do A , B 2. Viet phu-o-ng t r i n h du'cmg thang ( d ) di qua goc toa do 0 cat do t h i (C) y = - X tai 4 diem phan biet la 4 dinh cua 1 hinh chir nhat c6 dien tich bang ^ . Tu-ang tu- m > 0, ta c6: M - , / ^ ; - V 2 m , N^ , / ^ ; V 2 m ,Vm J Lo-i giSi Vi 0 la tam doi xu-ng ciia (C) nen ABMN la hinh binh hanh. Do do ABMN la 2x + l 1. Phu-cng t r i n h hoanh do giao diem: = x + m x + ( m - 3 ) x - m - l = 0, 32 x-1 hinh chu- nhat c6 dien tich bang — khi va chi khi x^l (*) . . , •OA = OM mk = l k =3 Vi phu-cng t r i n h (*) c6 A = m ^ - 2 m + 1 3 > 0 , V m e l nen phu-ang t r i n h (*) c6 'k = l 32=*- io»- 1 hoac • AM.AN = m = - 2 nghiem phan biet, tii-c du-crng thang da cho cat do t h i tai 2 diem phan biet y [ k +m=— 3 I 3 m =3 M,N v a i VmeM. Vay, ( d ) CO phu-ang t r i n h : y = - x , y = 3x . > / Goi M ( X I ; X I + m ) va N(x2;x2 + m) la toa do can t i m . 3 SiMN = 4 » i . M N . d [ l , ( d ) ] = 4 hay M N . d [ l , ( d ) ] = 8 (**)
cap ttfc giai 10 chuySn dg 10 difi'm thi mOn Toan - Nguygn Phu Khanh Cty TNHH MTV DWH Khang Vigt B a i tap tur l u y e n : A ' = (m + l ) ^ - ( 2 m + l ) 1. Cho ham so y ==-- i i i ^ c6 do t h i la (C) . T i m tat ca cac gia t r i tham so m G hai nghiem phan biet t > 0 S = 2 ( m + 1 ) > 0 m;tO va m > —. 2x-2 2 P=2m+1>0 de du-ang t h i n g ( d ) ; y = x + m c a t do thj (C) tai 2 diem phan biet A , B sao cho O A ^ + O B ^ - — . Vo-i - - < m ^ 0 t h i do t h i cua ham so cat true hoanh tai 4 diem phan biet c6 Hiro-ng dan giai hoanh dp theo thu-tir -yjt^'.-^Ji^'.yf^'.-Jt^ vai t^ > t 2 . .; ,!,. X+ 2 Phirang t r i n h hoanh do giao diem ciia (d) va ( C ) : =x+m Theo g i a t h i e t S^CK =^AC.d(K;AC) (3) v a i d(K;AC) = |yK ti+t2+27t7t7 = 16 A „ =:4m^ + 4m + 2 5 > 0 Ap dung dinh ly Vi-et cho phu-ang t r i n h (2), ta du'p'c: Vi s V m € l R nen (d) cat (C) tai 2 diem phan biet vo-i g(l) = 3^0 'm-7>0 ' 2(m + l ) + 2V2m + l =16c:>V2m + 1 = 7 - m •m = 4 . VmeR. m ^ - 1 6 m + 48 = 0 3. Tim tham so m sao cho do t h i Goi A { x i ; X ] . + m ) , B ( x 2 ; x 2 + m ) la toa dp giao diem ciia (d) va (C). (m + l)x + m-35 (C): y = x'+3x^ va ( H , J : y = eat nhau tai 2 diem phan biet. Theo dinh ly Vi-et, ta c6 x^ + X2 = ~ ~ Y ~ ' ^1 •'^2 = -{^ +1) • x+1 Hirang dan giki Ta c6: OA^ + OB^ = x^ + ( x j + mf +xl+ (x2 + mf Phu-o-ng t r i n h hoanh dp giao diem (C) va ( H ^ ) la: = 2 ( x i + X 2 ) ^ - 4 x i . X 2 + 2 m ( x i + X 2 ) + 2m^ 3 2 (m + l ) x + m - 3 5 3 „ 2 36 , 4 2 x''+3x''=-^^ '- ox^+3x'^+ =m+l (*) ^ 2m-3^ 2m-3 x+1 x+1 ^ ' = 2 + 4 ( m + l ) + 2m + 2 m ^ = - ( 4 m ^ + 2m + 17 ( C ) va ( H ^ ) eat nhau tai 2 diem phan biet k h i va chi k h i phu-o-ng t r i n h (*) eo Giathiet: O A ^ + 0 B 2 = — « i f 4 m 2 + 2 m + 17) 2 nghiem phan biet khae - 1 k h i do f ( x ) = m + l eo 2 giao diem. 2 2^ ! . Xet ham so: f ( x ) - x^ + 3x^ + — tren R \} , 2m^ + m-10:=0m = — hoac m = 2. 2 _36_ ^3 ( x + l ^ j ^ ^ Taeo: f'(x) = 3 ( x 2 + 2 x ) - Vay rn = hoac m = 2 la gia t r i can t l m . (x+ir (x+ir 2. C h o h a m s o y = x ' ^ - 2 ( m + l ) x ^ + 2 m + l c o d o t h i l a (Cn,) . T i m tat ca cac gia Va f ' ( x ) = 0 x = - 3 hoac x = : l trj tham so m e K de do thj ham so da cho cat true hoanh tai 4 diem phan biet lim f(x) = - o o , lim f(x) = +oo, lim f(x) = -co, lim f ( x ) = +oo A,B,C,D Ian lu-o-tco hoanh do X i , X 2 , X 3 , X 4 (x^ < X 2 < X 3 < X 4 ) sao cho tam giac Diravaobangbienthiensuyra m < - 1 9 hoac m > 2 1 . ; ACK CO dien tich bang 4 , biet K ( 3 ; - 2 ) . 1+x Hirang dan giii Cho ham so y = c6 do t h i la (C). T i m tham so m de du-o-ng t h i n g d^ : x ' ^ - 2 ( m + l ) x 2 + 2 m + l = 0 ( l ) . D a t t = x2 (t^O). y = x + 2m cat do t h i (C) tai hai diem phan biet A va B cung diem I tao phu'o-ng t r i n h ( l ) t r a t h a n h t ^ - 2 ( m + l ) t + 2m + l = 0 ( 2 ) . 1 1 thanh tam giac c6 dien ti'ch bang 1, vo-i I De do t h i ham so cat true hoanh tai 4 diem phan biet thi phu'ang t r i n h (2) c6 7-^
Ca'p ttfc giai 10 chuy6n dg 10 die'm thi m6n Toan - Nguyjn Phu Khanh Cty TNHH MTV DVVH Khang Vigt Hu-o-ng dan gi§i Hiro-ng d i n giai ( d ) : y = m ( x - l ) . Phu-ang trinh hoanh do giao diem cua ( d ) va ( C ) : m x ^ - ( 2 m + l ) x + m = 0, X 7 ^ l ( l ) ' S,^g = l r : > 1 6 m ' * + 3 2 m ^ + 1 2 m ^ - 4 m - 6 = 0 = > m = i Theo bai toan, ta c6 phu-ang trinh ( l ) c6 2 nghiem phan biet X i , X 2 thoa man X i < 1 < X2 => m > 0 . Gia su-x^, = X j , X N = X 2 . 5. T i m m de du-ang t h i n g d : y = x + m + 2 cat do thj (C.^,): y = +3x^ + m x - 1 Laico: X N + 2 X M = 3 = > m = 2 . tai 3 diem phan biet A,B,C sao cho BC = 4, X A = 1 . 9. Tim m^O de du-ang thang y = -x + m cat do t h i (C,„) ciia ham so: Hirang dan giai > h - y =ix'^-(m-2)x^+3(2m-3)x + m tai 3 diem phan biet A ( 0 ; m ) , B , C dong Phu-ang t r i n h hoanh do giao diem ( x - l ) ( x ^ + 4x + m + 3J = 0 thai O A la du-ang phan giac trong cua B O C trong A B O C . , Vai - 8 ; ^ m < l t h i d cat (Cfn) tai 3 diem phan biet , Hirang dan giai B C ^ = 2 ( x i + X 2 f - 8 x 1 X 2 = 8 ( l - m ) , BC = 4 = > m = - l . B(XB;-X|3+m), C(xc;-Xc+m) '.,>,.• 6. T i m m de diro-ng thang d : y = x - m cat do thi [C^]: y = x ^ + 3 x ^ + m x - 3 tai Vi O A la du-ang phan giac trong cua B O C trong A B O C nen , 3 diem phan biet X i , X 2 , X 3 sao cho bieu thu-c T = 2 ^ X 1 + X 2 + X 3 J + 3 x ^ X 2 X 3 - 5 A C _ O C ^2x1 _ V x c + ( x c - m f Xc-m_Xij-m A B ^ 0 B ^ ^ = ^^2^(^^_^^a" xc = X, datgia t r i nho nhat. Hirang dan giai 2xg.X(- - m(x|3 + X(^) = 0 => m^ - 14m + 10 = 0 Phu'ang t r i n h hoanh do giao diem cua d va (C^^): 10. Tim m de ham so : y = x - ( 2 m + 3)x + 2m - m + 9 j x - 2m + 3 m - 7 cat true (x + l ) ( x ^ + 2x + m - 3 ) = 0 , v a i m < 4 t h i T = 3 m ^ - 2 2 m + 44 hoanh tai 3 diem phan biet, trong do c6 2 diem c6 hoanh do Ian h a n 1 va 11 khoang each giu-a 2 diem nay la Ian nhat. 11 11 . „ 11 , , . 11 = 3 m -> — => m i n T = — khi m = - Hirang dan giSi 3 3 3 3 3 (x-1) x^-2(m + l ) x + 2m^-3m + 7 = 0 7. T i m m de du'o-ng thang d : y = m x - l cat do thi (C,,,): y = 2x,3 - 3 x + m tai 3 diem phan biet k{l;yp^), B,C sao cho M ( 2 ; 2 m - l ) nam t r o n g doan BC va Theo bai toan, phu-ang trinh x ^ - 2 ( m + l ) x + 2 m ^ - 3 m + 7 = 0 co 2 nghiem • MB-2MC •^ ^ • : 'i phan biet X i , X 2 thoa man 1 < X j < X2 2 < m m = 55 M« aanrM-: Hiro-ng dan giai 8. Tim m de dirang thang ( d ) qua diem A ( l ; 0 ) c6 he so goc m cat do thj (C): V m e R t h i ( d ) cat (C) tai 2 diem phan biet M,N va A ( - m ; 0 ) ? >: X+2 y =- — tai 2 diem phan biet M, N thuoc 2 nhanh ciia (C) sao cho AM = 4AN X i + m = 4 ( x 2 + m ) hoac X2 + m = 4 ( x 2 + m ) X ~ 1 A N + 2 A M = 0 va X M < X N . 7^
Ca'p te'c giSi 10 chuySn 6S 10 difi'm thi mOn Toan - Nguygn Phu Khanh Cty TNHH MTV DVVH Khang Vi?t 'i> C h i i dt 3: Diem thuoc d d thi. T H I : A ( - 3 ; - 3 ) va C(5;21). T i m tat ca cac diem M tiiuoc do tlij (C): y = f ( x ) , biet M tiioa man tinh chat d:- ^ va D € ( d ) ^ D(t;2-t) T ciio trirffc. V i d u 1. T i m t r e n do t l i i (C): y = - 3 x ^ + 1 , 2 diem M, N sao ciio MN = Asjl Taco: AD = (3 + t ; 5 - t ) , CD = (-5 + t ; - 1 9 - t ) va tiep tuyen tai do song song v a i nhau. DA.DC = 0 ABCD la hinh vuong k h i va chi khi L a i giki DA = DC Giasu- M ^ m . m ^ - 3 m ^ + N | n , n ^ - 3 n ^ + i j v a i m^^n la toa do tlioa de bai. •t = - l l V t= 5 "^1/ N2 N2 / x2 / ^2=>t = - l l tu-c D ( - l l ; 1 3 ) . Vi tiep tuyen tai M,N song song v 6 i nhau nen y ' ( m ) = y ' ( n ) {3 + t ) + ( 5 - t ) = ( t - 5 ) + ( - 1 9 - t f ^ ' hay 3 m ^ - 6 m = 3 n ^ - 6 n < = > ( m - n ) ( m + n - 2 ) = 0 =>n = 2 - m , m ; ^ l Vi AB = DC = (16;8) => B ( l 3 ; 5 ) H Vay, A ( - 3 ; - 3 ) , B ( l 3 ; 5 ) , C ( 5 ; 2 l ) , D ( - l l ; 1 3 ) la toa do can t i m . ' m' Hannij-a MN^ = : ( m - n ) ^ + ( m ^ - 3 m ^ + l - n ^ + 3 n ^ - l ) , r u t g p n ta du-gc T H 2 : A(5;21) va C ( - 3 ; - 3 ) tu-angtir. ' MN^ - 4 ( m - l f - 2 4 ( m - l ) ' ^ + 4 0 ( m - l f ,do n = 2 - m V i d u 3 . C h o h a m s 6 y = x ^ - 5 x ^ + 1 0 x - 8 , c 6 d o t h i (C) Ma MN = 4 V 2 suy ra 4 t ^ - 2 4 t ^ + 4 0 t = 32 v o i t = ( m - l ) ^ , t > 0 , giai ra du'gc 1. Goi A la diem thuoc (C), C la diem thuoc du-ang t h i n g d : x - 7 y + 25 = 0 va t = 4 , t u ' d a y c 6 ( m ; n ) = ( 3 ; - l ) hoac ( - 1 ; 3) r 17"! , ' la trung diem AC. T i m toa dp diem B c6 hoanh dp am sao cho tarn Vay, diem can t i m M ( 3 ; - l ) , N(-l;3). \ I j giac OAB vuong can tai A , K; V i d u 2. T i m toa do 2 diem B, D sao cho ABCD la hinh vuong, biet rang D la 2. Goi E , F theo thu- t g la giao diem cua du-ang t r o n ngoai tiep tii- giac OABC v a i diem nam tren du-ang t h i n g d : x + y - 2 = 0; i ( l ; 9 ) l a t r u n g d i e m AC; A va true hoanh, true tung (;E,F khac 0 ) . T i m tpa dp diem M tren du-ang tron 1 1 7 7 C la 2 diem nam tren do t h i y = - x ^ - - x ^ - - x + - . sao cho tam giac M E F c6 dien tich Ian nhat. 3 2 3 2 Lo-i g i a i L a i giai !• A(a;a^-5a2 + 1 0 a - 8 ) 6 { C ) , C(-25 + 7 c ; c ) e ( d ) f 1 3 1 2 7 7 1 3 1 2 7 7^ Goi A a ; - a - - a - - a + - , C c;-c - - C - - C + - 3 2 3 2 3 2 3 2 'a + (-25 + 7c) 1 la 2 toa do diem thupc do thj cua ham so. 2 2 la t r u n g diem AC a+c . I 22 (a^-5a^ + 1 0 a - 8 ) + c 7 = 1 2 2 " 2 I l a t r u n g d i e m AC: u fl 12 7 3 7> f l 3 1 2 7 7^ -a - - a - - a + - + -c - - C — C + -- 24-a U 2 3 2; 2 3 2) = 9 c=- a = 3^A{3;4) (a -3)(7a2 - 14a + 2?) = 0 ^ ^ = ^ ^ C(-4;3) a+ c= 2 J 1 i ( a + c)[(a + cf - 3ac] - i[(a + c f - 2ac] - ^ ( a + c) - 1 1 = 0 Goi B(xo;yo), X o < 0 . Tam giac OAB vuong can tai A k h i AB.OA = 0 AB = OA a+c= 2 a = -3 a=5 A(-3;-3) s hay hoac ngu-gc lai 3(xo-3) + 4(yo-4) =0 rxo=-l;yo=7 ac = - 1 5 " c=5 c = -3 [C(5;21) =25^U B(-l;7). (xo - 3 ) ' +(yo - 4 f =7;yo = 7 "
cap te'c giSi 10 chuySn OJ 10 cliS'm thi mOn Toan - Nguygn Phu KhAnh CtyTNHH MTV DVVH Khang Vi§t Goi H, K Ian lu-p-t la hinh chieu ciia B, C len du'crng thang y = - 2 , k h i do 7^ 2 2 ( X +11 25 2. Dirang t r o n ngoai tiep OABC : - + y " H ( b ; - 2 ) va K ( c , - 2 ) . Tu-gia t h i e t s u y ra E(-1;0), I F(0;7). 2) V 2y 2 De thay B ' A H + CAK = CAK + ACK = 90° =>B'AH = ACk suy ra AAHB = ACKA De thay, EF la dirang ki'nh duo-ng tron, nen tam giac MEF vuong tai M . fAH=CK, (canh huyen, goc nhon } hay 'T BH = AK 1.,^.,^ ME-^ + MF^ EF^ 25 S„EP=-ME.MF< =—= Y- Dang thii-c xay ra k h i va chi khi tam giac MEF vuong can tai M , k h i do toa dp M (i-br= 2+ - bc = 3c + 2 V bc = - c - 2 (x + 1) +y^=25 2+ ^ c-l M(-4;4) c-l b (2) thoa he: 2 f (X 1- ' 7^ M(3;3) + + y 3c + 2 I 2, 2) 8c + 4 * Voi bc = 3c + 2=> b = thay vao (2) ta du-gc c-l B a i tap tir l u y e n : ,., ^ 3c+ 2 1. T i m cac diem M tren do thj (C): y = x'* + 2x^ - 1 sao cho tiep tuyen cua (C) tai Suyra c'^+3c + 2 = 0 hoac 3c^+7c + 2 = 0 khongthoa o O , / 17^ -c-2 M vuong goc v a i du'crng t h i n g IM, v a i I 0;- * V a i be = -c - 2 => b = thay vao (2) ta du-gc c-l I' c+2 n Hirang dan giai Suyra c + c - 6 = 0 c = 2 hoac c = - 3 [ k h o n g t h o a o O ] Tiep tuyen d tai M(xo;yo) thuoc (C) c6 he so goc la y^ = 4 x o + 4 x o , phu-o'ng Vay, B ( - 2 ; - l ) , C ( 2 ; l ) hoac ngu'gc lai la toa dp can t i m . t r i n h c6 dang: y = | 4 X Q + 4 x o J ( x - X o ) + X Q + 2xo - 1 va c6 vecto phap tuyen 2x + l 3. Tim cac diem thuoc 2 nhanh khac nhau ciia (C): y = sao cho khoang x+1 n = (4xg+4xo;-l). each giij-a 2 diem do ngan nhat. f 25 Hirang dan giSi Du-crngthang A M c6 vecta chi phu-ang AM = X o ; X o + 2 x o 1 V 8 Goi A a;2- , B b;2- vo-i a < - l , b > - l la 2 diem Ian lu-ot thuoc a+1 b+1 Du'angthang d va AM vuong goc nhau khi n va AM cijng phu'o-ng voi nhau tu-c nhanh phai va nhanh trai cua do thi. ( T 4XQ+4XQ \f ^ Xo+2xo 25^ + X o =0Xo = 0 hoac 4 XQ + 1 /7 \ 7 XQ + 1 \ 33 + 1=0 (*) Datu = - l - a > 0 , v =l + b>0. 33 1 1 Dat t = XQ + 1 > 1 , phu-ang t r i n h (*) t r o thanh 4t + 1 = 0 , phu-ong A B 2 = ( u + v f + —+ — ={u+vy 1+- >4uv 1 + - 8 u vj (uv)^ (uvr t r i n h nay C O nghiem t = 2 thoa dieu kien t > l . Hay AB^ > 4 u v + — > 1 6 Vai t = 2 tu-c XQ = l < = > X o = - 1 hoac XQ = 1. ^fe'"-'*'^ I ' uv Vay.co 3 diem can t i m M ( - l ; 2 ) , M ( 0 ; - l ) , M(l;2). u =v U= V Bang thu-c xay ra k h i : • 4 •u=v=l 2 4uv = 2. T i m toa do 2 diem B, C thuoc 2 nhanh khac nhau c i i a d o t h i y = - sao cho uv X tam giac ABC vuong can tai A ( l ; - 2 ) . , Vay, A(0;1), B(-2;3) t h i m i n A B = 4 . = Hu-ang dan giai Tim tren do t h i (C) : y = - x ^ +3x c6 bao nhieu bp bon diem A,B,C,D sao cho 2^ ( 2] tu-giac ABCD la hinh vuong tam O(0;0). Xet B b;- , c c;- , b < 0 < c la 2 diem thuoc do t h i y = - . l b; X 29
Cafp tffc giSi 10 chuv6n 66 10 djgm thi mOn Toan - Nguygn PhCi Khanh Cty TNHH MTV DVVH Khang Vi?! Hiro-ng dan gi^i ^, Tim tren du-ang t h i n g y = 3x - 2 diem M sao cho tong khoang each tir M den GiasLT A ( a ; - a ^ + 3 a ) , B ( b ; - b ^ + 3 b ) vai a^^bva a , b > 0 . 2 diem circ t n ciia do t h i cua ham so y = x^ -3x^ + 2 la nho nhat. ), , Hiro-ng dan giM OAIOB JOA.OB = 0 ABCD la hinh vuong tarn O(0;0) Gia sir diem circ dai la A ( 0 ; 2 ) , diem ctrc tieu la B ( 2 ; - 2 ) . • tl . ,• OA=OB l0A=0B Ta thay, A,B nam ve 2 phia dirang thang y = 3 x - 2 . ab + a b ( a 2 - 3 ) ( b 2 - 3 ) = 0 E)e MA + MB nho nhat k h i 3 diem A,M,B thang hang va M nam t r o n g A B , a2+(a3-3a) =b2+(b3-3b) tire tpa do diem M la giao diem cua du-ang t h i n g A B : y = - 2 x + 2 va du-ang M 2^ [ a V - 3 ( a 2 + b2) + 10 = 0 (l) thang y = 3 x - 2 =>M 5'5 Bien doi va rut gon (*), ta du-gc : "a + b = 0 2x-l 7. Tim toa dp diem M tren do t h i (C) : y = sao cho khoSng each tir diem (a2+b2-3)'-aV+l = 0 (2) x+1 l ( - l ; 2) t a i tiep tuyen tai M eiia do thj (C) bang . ' ' «- ; Triro-ng ho-p 1 : a + b = 0 thay v^o (l), ta dirge : a'^ - 6a^ + 1 0 = 0 (3) . Ro rang 5 ,, , p h i r c n g t r i n h (3) khong c6 nghiem thirc v a i Va e M . Hu-ang dan gi^i 'E Tru-o-ng h a p 2: Dat u = a^ + b ^ v = a^.b^. Gpi M X o ; 2 - XQ+I la tpa dp diem can t i m , tiep tuyen tai M c6 phu-ang t r i n h : v - 3 u + 10 = 0 fv = 3 u - 1 0 K h i d o h e (1), (2) t r a t h a n h : . 3 3 u ^ - 9 u + 20 = 0 y-2 + • ^' ^^ [(u-3) - v+ l =0 XQ + I (XO + 1 ) ' { x - X o ) h a y 3 ( x - X o ) - ( x o + l f ( y - 2 ) - 3 ( x o + l ) = 0 ( t ) Giai he, ta du-ac u = 4,v = 2 hoac u = v = 5 . u =4 a2+b2=4 a = 72-V2 a= V2 + \/2 Khoang each tir 1 den tiep tuyen ( t ) la d = 3 ( - l - X o ) - 3 ( x o + l ) | 6xo + l hoac v =2' b-V2-72 >/9+(xo + l ) ' ^/9+(xo + l)' b= V2 + V2 d = SsflO 6 ^ 7 J ^ | x p + i | ^ ^ 9 ^ ^ X o + l)'^ ^ t ^ - 1 0 t + 9 = 0 •'•'A u =5 \a^+h^=5 a= a= 5 VlO hoac v = 5' alb2=5 (5-75 v a i t = ( x o + l ) ^ > 0 t = l hoac t = 9 5 + N/5 b = b = Vi vai t r o A,B nhir nhau nen tren (C) c6 hai bo bon diem A,B,C,D sao cho *^ Chii de 4: Tinh don di^u ciia ham so. ABCD la hinh vuong CO tarn O(0;0). 1- Dieu kien can de h^m so d e n dieu: 5. T i m tren do thj (C) cua ham so y = + 3x - 2 cap diem doi xirng nhau qua goc GiS su- ham so f c6 dao ham tren khoang I y: ' • Neu ham so f dong bien tren khoang I t h i f ' ( x ) > 0 v a i mpi x 6 1 ; ',]•> • . t p a d o I (2; 1 8 ) . • Neu ham so f nghich bien tren khoang I thi f ( x ) < 0 v a i mpi x e I . ;} Hirang d§n giSi 2- Diliu kien du de h^m so d e n dieu: Gpi M ( x i ; y i ) , N ( x 2 ; y 2 ) latoadocantim.Tirgiathiet.tasuyra: Gia su- I la mot khoang hoac nu-a khoang hoac mot doan, f la ham so lien tuc yi+y2=36 tren I va c6 dao ham tai mpi diem trong ciia I (tire la diem thupc 1 nhirng Vay, M ( l ; 2 ) , N(3;34) khong phai dau m i i t ciia I ] . Khi do: • J^eu f ' ( x ) > 0 v a i m p i x e 1 t h i ham so f dong bien tren khoang I ; 30
Cty TNHH MTV DWH Khang Vi§i cap tflc gi^i 10 chuvfin 10 diSm thi mfln Toan - Nguyen Phii Khanh ^ Xi 0 , 2 a < X i + X20, ( x i -P)(X2 -P) > 0. • ^' ' ' ' ' ' ' ' ' ' ' ' •' ' ' Ham so da cho xac dinh tren D = IR \, do do no du-o-c xac dinh tren khoang • Mfrrong: Nhan thay, v a i - 4 < m < 4 t h i ham so da cho luon dong bien tren khoang {l;+oo). NghTa la ham so cung dong bien bat ke khoang hoac nu-a khoang hay Taco:y' = . ; doan nao thupc ( l ; + o o ) . Nhu- vay, ham so hien nhien dong bien ( d a n dieu (x + 3) tang) bat k i tren khoang (1;2) hoac doan [3;5] ,,,,,0, frtt s ^ Ham so da cho dong bien tren khoang (l;+oo) k h i va chi k h i y ' > 0 v o l Qua do, bai toan c6 the yeu cau: "Tim m de ham so: y = ^L±5itE_±^ dong X+ o V x € ( l ; + o o ) t u x la x ^ + 6 x - r 9 - m ^ > 0 , V x e ( l ; + o o ) (vi (x + 3 f > 0 , Vx>l] bien tren doan [2;3]".... De hieu ky h a n van de nay, ban doc lam bai toan sau: h a y ( x + 3) > m v a i V x e ( l ; + o o ) . .p,^.,,,!! "Tim dieu kien tham so m sao cho ham so y = 4mx^ - 6x^ + (2m - l ) x + 1 tang Xet g ( x ) = (x + 3)^ t r e n k h o a n g ( l ; + c o ) v a g'(x) = 2(x + 3) v a i x > l =>x + 3 > 4 t r e n k h o a n g (0;2)". " i i ^ < s ^^oa .^ tii-c g ' ( x ) > 8 > 0 v o i V x € ( l ; + o o ) . ,, g ( x ) dong bien tren khoang (l;+=o) va l i m g { x ) = 1 6 , l i m g ( x ) = +oo. V i d u Z . T i m m d e h a m s o : y = x ^ - m x ^ + ( m + 3 6 ) x - 5 nghich bien tren khoang Khi do m^ < ( x + 3 ) ^ V x 6 ( l ; + o o ) m^ < 16 hay - 4 < m < 4 . CO do dai bang 4 ^ . ^ ^ .^^ Chiiyl: !•< CI : r i ' L M gi^i * Wdmso y = f ( x , m ) t d n o t r e r ? R o y ' > 0 Vx€lRmin y ' > 0 . Ham so da cho xac djnh tren R . * Ham so y = f ( x , m ) gidm tren R o y ' < 0 V x € K max y ' < 0 . Taco: y ' = 3 x ^ - 2 m x + m + 36 va A' = m ^ - 3 m - 1 0 8 Chuy2: Dot f ( x ) = ax^+ bx + c ( a ^ t O ) . 1 i iu'ty" De thay ay. = 3 > 0, do do ham so da cho khong nghich bien tren R . • f ( x ) = 0 CO hai nghiem x^,X2 thoa man : X i < a < X 2 . Dqt t = x - a , khi do Neu m < - 9 hoac m > 1 2 tii-c A ' > 0 t h i y ' = 0 co 2 nghiem phan biet X j ; X2. g ( t ) = f ( t + a ) . Bai toan tra thanh g ( t ) = 0 c6 hai nghiem trdi ddu tiic Lap bang xet dau, ta thay y ' < 0 v a i x e ( x i ; x 2 ) suy ra ham so nghjch bien v a i ti
Ca'p tOc giai 10 chuygn dg 10 digm thi mSn Toan - NguySn Phu Khjnh Cty TNHH MTV DVVH Khang Vi§t Bai tap tir luyen: -x^+2x >m, VXG(0;1). 1. Tim a de ham so y = + ax^ + 4x + 3 dong bien tren K . 4x + l 3 - x ^ +2x ,. Hirang d i n gi^i X e t h a m so g ( x ) = lien tuc tren khoang ( 0 ; l ) . 4x + l Ham so da cho xac dinh tren R . Ta c6 y ' = x^+2ax + 4 va CO A' = a ^ - 4 -4x^-2x + 2 Taco: g'(x) = in,;- r ^.• v. CAch 1: Ham so da cho dong bien tren M y ' > 0, Vx e K nghla la ta luon c6: (4x + l f A' = a ^ - 4 < 0 -2 0 , x ^ - 2 . Ham so y dong 2 khong thoa man yeu eau bai t o a n . ham tren cac khoang (a;xo) va ( x g i b ) . Khi do : Vay ham so y dong bien tren R k h i va ehi k h i - 2 < a < 2. f'(xo)0, VxeMo * y'0 a
Ca'p tO'c giai 10 chuySn d l 10 djgm thi m6n Toan - NguySn Phu Khanh Cty TNHH MTV DWH Khang Vi$t Lo-i giai Neu m < 0 t h i 2x^ + m = 0 c6 hai nghiem phan biet khac 0 , do do ham so da Ham so da cho xac djnh va lien tuc tren khoang (-oo;-m)i^(-m;+oo). cho CO 3 eye t r i . Vay, m < 0 ham so da eho eo 3 eye t r i 1 , „ 1 Ta c6: y ' = l - - va y =• , 2 ^ ^ 2 ^ (x + m)^ (x + m) m , m ^ 3m m ^ 3m 0;6 , B va C —;o Ham so c6 dao ham tai mpi diem tren khoang xac djnh, nen ham so dat eye tieu V 2 4 tai x = l k h i thoa man: 2 ^ _ ^ 2\ m m m m 1. C a c h l : AB = va AC = Dieu kien can: y ' ( l ) = 0l • ^- = 0m = 0 ; m = 2 (1 + m) Dieu kien du: Tam giac ABC vuong tai A khi A B 1 AC hay AB.AC = 0 \\ m = 0 = > y " ( l ) = l > 0 = > x = l la diem eye tieu. m m m"^ m 2 ^ ( 7 , ^ =0 « ^ = 0, ' V m = 2 = > y " ( 1 ) = - l < 0 = > x = l l a diem eye dai. T T 2 V / Vay m = 0 thoa yeu eau bai toan. phu-ang t r i n h nay eo nghiem m = - 2 ( t h o a m < 0 ) hoac m = 0 khong thoa. NMn xet: De y dinh ly 3 ehi phat bieu khi y " ( l ) 0. Vay, m = - 2 thoa de bai. ' fy'(l) = 0 Cach2:Goi I l a t r u n g d i e m BC; do tam giac ABC vuong can tai A Neu t r i n h bay ham so dat eye tieu tai x - 1 • thi l a i giai chu-a ehinh y"(i)>o 2 - A . BC . m m , m"* m „ m m nen AI = — tu-c — hay — + — = 0 — 1 = 0=> m = - 2 2 4 [y'(i)=o 2 16 2 2 T xac. Nhu- vay, de ap dung du-o-e he tacan khangdjnh y " { l ) > 0 . y"(l)>0 2. Dien tich tam giac ABC bang 32 khi va ehi khi ^ ^ = 3 2 hay J - - . — = 1 6 (*) Chiiy: 2 V 2 4 * Ham so f (xdc dinh tren D)c6 cite tri BXg € D thoa man hai dieu kien sau: Dat u = J - Y => m = -2u'^, khi do phu-ang trinh (*) tree thanh i) Tai dao ham cua ham so tai Xg phdi triet tieu hogc ham sokhong c6 dao ham tai Xr u ^ = 3 2 = 2^=^u = 2 tu-e m = - 2 ( 2 ) = - 8 thoa man. j;*^^ '' ii) f'(x) phdidoi ddu qua diem XQ hogc f"{xo)^0. Neu f ' ( x ) la mot tam thitc bgc hai hogc triet tieu vd cung ddu v&i mot tam thuc m 3. Ta c6:0A = 6 - va BC = 2 J - Y . Dien tieh tu- giac OABC bang 52 k h i va ehi bgc hai thi ham c6 cue tri
Ca'p tOc giai 10 chuySn dg 10 digm thi mOn Toan - Nguyin Phu Khanh Cty TNHH MTV DVVH Khang Vi$t ( 4 Goi A 2; 2m;—m^ - 4 m ^ la 2 diem cu-c t r i va Datt = . J ^ , t > 0 , k h i d 6 ( * ) tro-thanh | l 2 - 4 t ^ | . t = 104 (**) 4m , B 3 3 THI: 0 < t < ^ = > 1 2 - 4 t ' ^ > 0 , p h u - a n g t r i n h (**) tro-thanh ( l 2 - 4 t ' ^ ) . t = 104. 2 3 .„2 ~ 2^ m + l ; - m ' ' - 2 m - 2 m + - la tpa do trung diem ciia AB. 3 3 De dang chirng minh du-gc o l e 2 x - 3 y = 0 m ( m ^ - 3 m - 4 ) = 0 m 6 { - 4 ; - l ; 0 } f ( t ) = ( l 2 - 4 t ^ ) . t - 1 0 4 < 0 vo-i mgi t € ( 0 ; ^ . ft'; ol.. 'is, ^ , t a c 6 1. T i m m d e h a m s o y = ( m + 2 ) x ^ + 3 x ^ + m x - 5 eo eye dai, eye tieu c6 hoanh do f'(t) = 2 0 t ^ - 1 2 = 4(5t'^-3) la cac so du-ang. Hiro-ng dan giai Vi t > ^ nen S t " ^ - 3 > 1 2 = > f ' ( t ) > 0 , moi t > ^ , s u y r a f ( t ) la ham so dong Ham so da cho eo cac diem eye dai, eye tieu eo hoanh do la cac so du-o-ng khi va bien tren khoang (>/3;+=c] ; han nOa lim f(t) 0 =>dothj h i k h i p h u a n g t r i n h : y ' = 3(m + 2 ) x ^ + 6 x + m = 0 eo 2 nghiem du-o-ng phan a = ( m + 2)?i0 ham so f ( t ) cat true hoanh tai 1 giao diem t e ( t / 3 ; + o o ) va f ( 2 ) = 0 , do do A' = 9 - 3 m ( m + 2 ) > 0 biet, nghTa la ta luon eo: P = m >0 phu-o-ng t r i n h ( 4 t ' ^ - 1 2 ) . t = 104 c6 nghiem duy nhat t = 2 tu-e hay 3 ( m + 2) V m = -8. m +2 4. Tu-giae ABOC la hinh binh hanh k h i va chi k h i BA = OC (*) A' = - m ' ^ - 2 m + 3 > 0 -3 < m < 1 2^ m m _rn g_3m_ Ta c6: BA = va 0C = mBH.AC = 0 (*). \CB=HrI A C ^^
cap toe giai 10 chuyen dg 10 diem thi mon Toan - Nguyjn Phu Khanh Cty TNHH MTV D W H Khang Vi§t 5. T i m cac gia t r j cua tham so m de do thj (Cn,) »ua ham so Ta c6: BH = f V 2 m ; - 4 m ^ + 4 m + — , AC = (^y2m;4m^j. 4 j y = - x ^ + 3 x ^ + 3 m ( m + 2)x + l c6 2 diem eye t r j A,B ma do dai AB = 2\/5 . 31 Khi do (*) 2m + 4 m ' - 4 m ^ + 4 m + =0 hay 8 m ^ - - 8 m ^ - y m - l = 0, Hu-6-ng dan gi^i Taco: y ' = - 3 x ^ + 6x + 3 m ( m + 2 ) . phu'cng t r i n h CO nghiem m = 2 thoa m > 0 i v (! 3 V ' " 3. Gia su-do t h i y = x " ^ - 2 ( m ^ + l ) x ^ + 3 c6 3 circ t r i A, B, C . T i m m de diro-ng Do t h i (Cn,) ciia ham so eo 2 diem eye t r i A,B k h i va chi k h i y ' = 0 c6 2 t r o n noi tiep tam giac ABC c6 ban kinh bang 1 . nghiem phan biet Xi,X2 A'=:9(m + l ) ^ > O c : > m : ? i - l . Hirang dan giai Tathay, y = - ( x - l ) y ' + 2 ( m + l ) ^ x + ( m + l f va do y ' ( x i ) = y ' { x 2 ) = 0 Ham so da cho xac dinh tren K . , , •3 Taco: y ' = 4x x ^ - m ^ n e n s u y r a y ( x i ) = 2(m + l ) ^ x ^ + ( m + l ) ^ , y ( x 2 ) = 2 ( m + l ) ^ X 2 + ( m + l ) ^ De thay, V m e R t h i y ' = 0 c6 3 nghiem x = 0 hoac x = -Vm^ + l hoac Taco: A | x i ; 2 ( m + l f X j + ( m + l ) ^ j , B|x2;2(m + l ) ^ X 2 + ( m + l ) ^ j x = Vm^ + 1 nen do thi ham so c6 3 eye trj. Suy ra AB = ^(x2 - x i f + 4 ( m +1)'^ (x2 - x^ )^ Giasu-A{0;3), sf-Vm^+ l ; 3 - ( m 2 , cfVm^ + l ; 3 - ( m ^ +1)^ -i2 = ^{^2-^\f 4(m + l ) \ (x2+xi)^-4xi.Xi 4(m + l)'* + l Ta c6: AB = AC = .^(m^ +1)'^ + m^ + 1 , BC = 2 V m ^ + l , I la trung diem BC = j 4 + 4 m ( m + 2)] 4(m + l ) +1 =2J(m + l) 4(m + l ) +1 =>AI = (m^ + l ) ^ AB = 2 V 5 c ^ ^ ( m + l f 4 ( m + l)'^ + l = 7 5 « ( m + l ) ^ 4 ( m + l)'^ + l = 5. Dien tich tam giac ABC: i.BC.AI = i ( A B +AC + BC)r v a i r la ban kinh dirang fiat t = ( m + l ) ^ > 0 = i > t ( 4 t ^ + l ] = 5 < = > 4 t ^ + t - 5 = 0 t r o n noi tiep tam giac ABC . m^ + 1 Vm^ + 1 o ( t - l ) r ( t +1)^ + 1 = 0t = l • r = l = l h a y ( m ^ + l ) = l + J(m^+l) + 1 (*) |(m^ + lj'^ + m^+l + Vm^ + l Vdi t = l = > ( m + l f = lm = - 2 hoac m = 0 thoa de bai. \ * ' 6. C h o h a m s o : y = x ^ + 3 x ^ - ( m + l ) x + 2 c6 do thj la { C ^ ) . T i m m de ham so c6 Dat t = m^+l. eye dai, eye tieu dong thai du-ang thang noi eye dai, eye tieu ciia ham so tao t^-l>0 Phu-ang t r l i h (*) v i e t l a i : t^ = l + Vl + t^ ^ t =2 v a i dirang thang y = 2x + 3 mot goe 45°. t^-l) = l + t^ Hirang dan giSi m > - 4 t h i do thj cua h^m so eo eye dai, eye tieu. Vai t = 2 tu-c m^ + l = 2 m = ± l . 1 2 1 4 . Gja SU" do t h i y = x''^ - 2mx^ + m c6 3 circ t r i A, B, C . T i m m de dirang tron Khi do y = - ( x + l ) y ' - - ( m + 4)x + - ( m + 7 ) . Do cac hoanh do eye t r j la nghiem 3 3 3 |pi tiep t a m giac ABC c6 ban kinh bang 1 . Hirang dan giai ciia y' = 0 nen cac diem eye t r i eo toa do thoa man du-ang thing m S 0 thi do t h i ham so c6 3 circ t r i A ( 0 ; m ) , B^-x/m,-m^ + m j , c|\/m;-m^ + m j . 3 3 Giai n h i r b a i ^ , ta t i m dirac m = 2. 19 9 Tir gia thiet, suy ra m = hoac m = - ,„ An 6 2 ' ••' '• /II