2 PHNG PHP GII MT S DNG BI TPKHO ST HM S TRONG K THI TSH Phn mt: Cc bi ton lin quan n im cc i cc tiu A) Cc i cc tiu hm s bc 3:3 2ax y bx cx d = + + +* ) iu kin hm s c cc i cc tiu l: y=0 c 2 nghim phn bit * ) Honh im cc i cc tiu k hiu l 1 2, x xkhi 1 2, x x l 2 nghim ca phng trnhy=0 * ) tnh tung im cc i cc tiu ta nn dng phng php tch o hm tnh phng trnhng thng i qua im cc i cc tiu+ C s ca phng php ny l: nu hm s bc 3 t cc i cc tiu ti 1 2, x xth 1 2'( ) '( ) 0 f x f x = =+ Phn tch'( ). ( ) ( ) y f x px hx = + . T ta suy ra ti 1 2, x xth 1 1 2 2( ); ( ) ( ) y hx y hx y hx = = =l ng thng i qua im cc i cc tiu + K hiu k l h s gc ca ng thng i qua im cc i cc tiu * ) Cc cu hi thng gp lin quan n im cc i cc tiu hm s bc 3 l: 1) Tm iu kin ng thng i qua im cc i cc tiu ca hm s song songvi ng thng y=ax+b + iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu+ Gii iu kin k=a 2) Tm iu kin ng thng i qua im cc i cc tiu vung gc vi ng thng y=ax+b+iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu+ Gii iu kin k=1aV d 1)Tm m ( )3 27 3 f x x mx x = + + +c ng thng i qua cc i, cc tiu vung gc vi ng thng y=3x-7. Gii: hm s c cc i, cc tiu2'( ) 3 2 7 0 f x x mx = + + =c 2 nghim phn bit 221 0 21 m m ' A= > > . Thc hin php chiaf(x) cho f(x) ta c: ( ) ( )21 1 2 7. 21 33 9 9 9mf x x m f x m x ('( = + + + ( . Vi21 m >th f(x)=0 c 2 nghim x1, x2 phn bit v hm s f(x) t cc tr ti x1,x2. www.VNMATH.com3 Do 12( ) 0( ) 0f xf x' = ' = nn ( )( )21 122 22 7(21 ) 39 92 7(21 ) 39 9mf x m xmf x m x= + = + . Suy ra ng thng i qua C, CT c phng trnh( ) ( )22 7: 21 39 9my m x A = + Ta c( )( )2 2 221 21 213 72 3 4521 .3 1 219 2 2m m my xm m m > > > A = = = = 3 102m = 3) Tm iu kin ng thng i qua im cc i cc tiu to vi trc Ox mt gc+iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu+ Gii iu kintan k =V d 1) Cho hm s2 32 3+ = mx x x y (1) vi m l tham s thc Tmm hm s (1) c cc tr, ng thi ng thng i qua hai im cc tr ca th hm s to vi hai trc ta mt tam gic cn. Gii: Hm s c cc tr khi v ch khiy = 0 c 2 nghim phn bit ' 9 3 0 3 m m A= + > > 3 21 23 2 ( 1). ' ( 2) 23 3 3m my x x mx x y x = + = + + ng thng qua hai im cc tr ca th hm s c phng trnh 32 ) 232(mxmy + =ng thng ny ct 2 trc Ox v Oy ln lt tai |.|

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\|+36; 0 , 0 ;) 3 ( 26 mBmmATam gic OAB cn khi v ch khiOA OB =6 62( 3) 39 36; ;2 2m mmm m m =+ = = = Vi m = 6th O B A so vi iu kin ta nhn 23 = mCh : Ta c th gii bi ton theo cch: ng thng qua C, CT to vi 2 trc ta tam gic cn nn h s gc ca ng thng l 9( )22tan 45 1 2 13 3( )2m Lmkm TM

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www.VNMATH.com4 4) Tm iu kin ng thng i qua im cc i cc tiu to vi ng thng y=ax+b mt gc+iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu+ Gii iu kintan1k aka=+ V d ) Tm m ( )3 2 23( 1) (2 3 2) ( 1) f x x m x m m x m m = + + c ng thng i qua C, CT to vi 154y x= +mt gc 450. Gii: Gi h s gc ca ng thng i qua C, CT l k, khi t iu kin bi ton suy ra: 01 1 5 311 4 4 4 4 445 1 11 1 3 5 4 41 . 14 4 4 4 4k kk kktg kk kk k| | + = = | \ .= = + = | | + + = = | \ .3553kk

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Hm s c C, CT 2 2( ) 3 6( 1) (2 3 2) 0 f x x m x m m ' = + + =c 2 nghim phn bit23 5 3 53( 3 1) 02 2m m m m| | | | +' A= + > < > || ||\ . \ . (*) Thc hin php chia f(x) cho) f(x ta c | | ( )| |21 2( ) ( 1) . ( ) 3 1 ( 1)3 3f x x m f x m m x m ' = + vi m tho mn iu kin (*) th f(x)=0 c 2 nghim phn bit x1, x2 v hm s t ccc tr ti x1,x2. Do 12( ) 0( ) 0f xf x' =

' = nn ( ) ( )( ) ( ) ( )21 122 22( 3 1) 1323 1 13f x m m x mf x m m x m ( = +

( = +

Suy ra ng thng i qua C, CT c phng trnh( ) ( ) ( )22: 3 1 13y m m x m ( A = + Ta c( ) A to vi 154y x= +gc 450 ( )223 1 13m m + = kt hp vi iu kin (*) ta c 3 152m=5) Tm iu kin ng thng i qua im cc i cc tiu ct hai trc Ox, Oy ti A,B sao cho tam gic OAB c din tch cho trc +iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu+ Tm cc giao im vi cc trc to : Vi trc Ox:Gii y=0 tm x.Vi trc Oy gii x=0 tm y.+ /1.2MAB MABS d AB =T tnh to A, B sau gii iu kin theo gi thitwww.VNMATH.com5 V d 1) Tm m ng thng qua cc i cc tiu ca th hm s 33 2 y x mx = +ct ng trn tm I(1;1) bn knh bng 1 ti A,B m din tch tam gic IAB ln nht. Gii: C: 2' 3 3 y x m = c 2 nghim phn bit khi0 m > . Khi ta hai im cc tr ca th hm s l ( ) ( ); 2 2 , ; 2 2 M m m x N m m x +- Phng trnh ng thng MN l:2 2 0 mx y + =- ng thng MN ct ng trn tm I ti A,B m tam gic IAB c 2. . .sin 1IABS IA IB AIB = s , du bng xy ra khi 090 AIB = , lc khong cch t I n MN bng 12 Do vy ta c pt:( )22 11 1 3 3, 1 ; 12 2 2 24 1md I MN m mm= = = + = + V d 2) Cho hm s 33 2 y x mx = + Tm cc gi tr ca m th hm s c 2 im cc tr A, B sao cho tam gic IAB c din tch bng18 , trong ( ) 1;1 ILi gii: Ta c ( )2 2' 3 3 3 y x m x m = = . hm s c C v CT 0 m >Gi A, B l 2 cc tr th ( ) ( ); 2 2 ; ; 2 2 A m m m B m m m + PT ng thng i qua AB l: ( ) ( )42 2 2 22m my m m x m y mxm + = + = Khong cch t I n ng thng AB l( )22 1;4 1md I ABm=+ di on 34 16 AB m m = +M din tch tam gic IAB l 322 1118 4 16 1824 1mS m mm= + =+ ( )( ) ( ) ( )( )( )2 23 23 2 24 16 2 1 4 1 4.18 2 1 184 4 18 0 2 4 4 9 0 2m m m m m mm m m m m m m + = + = + = + + = = 6) Tm iu kin im cc i cc tiu cch u im M cho trc: +iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu ( Da vo phng trnh tnh gi tr 1 2; y y ) + Gi s im im cc i cc tiu l A, B th iu kin l MA=MB 7) iu kin im cc i cc tiu i xng nhau qua ng thng y=ax+b +iu kin l : y=0 c 2 nghim phn bit + Vit phng trnh ng thng i qua im cc i cc tiu ( Da vo phng trnh tnh gi tr 1 2; y y ) + Gi s im im cc i cc tiu l A, B th iu kin l: ng thng i qua im cc i cc tiu vung gc vi ng thng y=ax+b v trung im ca AB thuc ng thng y=ax+b www.VNMATH.com6 V d 1) Tm m hm s 3 2 2( ) 3 f x x x mx m = + +c C v CT i xng nhau qua ( )1 5:2 2y x A = . Gii: Hm s c C, CT( )3 26 0 f x x x m ' = + =c 2 nghim phn bit 2 29 3 0 3 3 m m m ' A= > > < . thc hin php chia f(x) cho f(x) ta c:( ) ( )221 2( ) 1 ( ) 33 3 3mf x x f x m x m ' = + + +vi3 m nn f(x)=0 c 2 nghim phn bit x1, x2 v hm s t cc tr ti x1, x2 vi cc im cc tr l .( ) ( )1 1 2 2; , ; Ax y Bx yThc hin php chia f(x) cho f(x) ta c:( ) ( )21 2 2( ) . ( ) 1 13 3 3f x x m f x m x m| |' = + + + |\ . Do 12( ) 0( ) 0f xf x' = ' = nn ( )( )21 1 122 2 22 2( ) 1 13 32 2( ) 1 13 3y f x m x my f x m x m | |= = + + + | \ . | |= = + + + |\ . Ta c( ) ( ) ( ) ( ) ( )22 2 2 22 22 1 2 1 2 1 2 1419AB x x y y x x m x x = + = + +

( ) ( )( )2222 1 1 222 244 1 194 4 2 134 4 1 1 4 19 9 3x x x x mm m AB ( (= + + + ( ( | | ( = + + + > + > | ( \ . Min AB=2 133 xy ram=0 9) Tm iu kin honh im cc i cc tiu tho mn mt h thc cho trc +iu kin l : y=0 c 2 nghim phn bit + Phn tch h rhc p dng nh l vit( 1 2, x xl hai nghim ca phng trnh y=0 V d 1) Tm m hm s 3 21( ) 13f x x mx mx = + t cc tr ti x1, x2 tho mn 1 28 x x >www.VNMATH.com8 Gii: Hm s c C, CT2( ) 2 0 f x x mx m ' = + =c 2 nghim phn bit ( ) ( )20 0 1 m m m m ' A= > < > vi iu kin ny th f(x)=0 c 2 nghim phn bit x1, x2 v hm s t cc tr ti x1, x2 vi x1+x2=2m v x1x2=m. Ta c BPT:21 2 1 28 64 x x x x > >( )22 21 2 1 24 4 4 64 16 01 65 1 652 2x x x x m m m mm m + = > >| | | | s > || ||\ . \ . tho mn iu kin( ) ( ) 0 1 m m < > V d 2) Cho hm s1 32 3+ + = mx x x yTm m hm s c cc i cc tiu v khong cch t im)411;21( In ng thng ni im cc i v cc tiu l ln nht Gii: Ta cm x x y + = 6 3 '2. Hm s c cc i cc tiu khi y=0 c 2 nghim phn bit 3 0 ' < > A m (0,25 im) - Chia a thc y cho y ta c13) 232( )313( ' + + + =mxm xy y. Lp lun suy ra ng thng i qua cc i cc tiu lA13) 232( + + =mxmy . D dng tm c im c nh m ng thng cc i cc tiu lun i qua l) 2 ;21( A(0,25 im) - H s gc ca ng thng IA l 43= k . H IH vung gc viA ta c 45/= s =AIA d IHI ng thc xy ra khiA IA(0,25 im) - Suy ra 34 1232 = = km1 = m(0,25 im) V d 3) Cho hm s 3 2 2 33 3( 1) 4 1 y x mx m x m m = + + (C) Tm m hm s c hai cc tr l A, B cng vi gc O to thnh tam gic vung ti O Gii:iu kin hm s c 2 cc tr l y=0 c hai nghim phn bit: 2 21' 3 6 3( 1) ' 9 01x my x mx mx m= + = + A= > = (0,25 im) Ta c 1 1'( ) 2 3 13 3y y x m x m = + Gi A, B l 2 im cc tr th ( 1; 3); ( 1; 1) A m m Bm m + + (0,25 im) Suy ra 21( 1; 3); ( 1; 1) 2 2 4 02mOA m m OBm m m mm= + + = = (0, 25 im) Kt lun: C hai gi tr ca m cn tm l m=-1 hoc m=2 www.VNMATH.com9 V d 4) Tm cc gi tr ca m hm s ( )3 2 21 1. 33 2y x m x m x = + c cc i 1x , cc tiu 2xng thi 1 2; x xl di cc cnh gc vung ca 1 tam gic vung c di cnh huyn bng 52. Gii: Cch 1: Min xc nh:D R =c 2 2 2 2' 3; ' 0 3 0 y x mx m y x mx m = + = + =Hm s c cc i 1x , cc tiu 2xtha mn yu cu bi ton khi v ch khi PT' 0 y =c 2 nghim dng phn bit, trit tiu v i du qua 2 nghim . 220 4 0 2 20 0 0 3 203 3 3 0m mS m m mPm m m A > > < < > > > < < >< v > > (*) Theo Viet ta c: 1 221 23x x mx x m+ = = . M ( )( )22 2 2 21 2 1 2 1 25 142 4 5 2 4 3 52 2x x x x x x m m m + = + = = = i chiu K(*) ta c gi tr 142m =tha yu cu bi ton. B) Cc i cc tiuhm s bc bn: 4 2ax y bx c = + + .*) iu kin hm s bc bn c 3 cc i cc tiu l y=0 c 3 nghim phn bit + Ta thy hm s bc bn th y=0 lun c mt nghim x=0, y=0 c 3 nghim phn bit sau khi tnh o hm ta cn tm iu kin phn phng trnhbc 2 cn lic 2 nghim phn bit khc khng. VD: 4 22 2 2 y x mx = + +th 3 2' 4 4 ' 0 0 y x mx y x x m = + = = v = iu kin l m0 Vi m>0 th f(x)=0 ( )( )( )4 21424 23; 20 0; 2; 2x m B mm m mx A m mx m C mm m m

= +

= +

= +

Suy ra BBT ca hm s y=f(x) AABC u 2 22 200mmAB AC AB ACAB BCAB BC> > = = ==

( )4 4 3340033 04mmm m m m mmmm m m> > + = + = = + = V d 2) Cho hm s 4 2 22 2 4 y x mx m = + , m l tham s thc. Xc nh m hm s c 3 cc tr to thnh 1 tam gic c din tch bng 1. Gii: Mx:D R = . C 3' 4 4 y x mx = 3 2' 0 4 4 0 0 y x mx x x m = = = v = . Hm s c 3 cc tr0 m >(*) Gi ( ) ( ) ( )2 2 20; 2 4 , ; 4 , ; 4 A m B mm C mm l 3 im cc tr Nhn xt thy B,C i xng qua Oy v A thuc Oy nn tam gic ABC cn ti A KAH BC c 21. 2 2 2 2 . 12ABC B A BS AH BC y y x m m mA= = = = . i chiu vi iu kin (*) c1 m =l gi tr cn tm. V d 3) Cho hm s ( )4 2 22 1 1. y x m x m = + + Tm m hm s cho c 3 im cc tr v ba im cc tr ny to thnh mt tam gic c din tch ln nht. Gii: ( )3 2 2 2' 4 4 1 0 0, 1 y x x m x x m = = = = hm s c 3 cc tr1 1 m < < . Khi ta im cc i l( ) 0;1 A m + , ta hai im cc tiu l ( ) ( )2 2 2 21 ; 1 , 1 ; 1 B m m C m m din tch tam gic ABC l( )( )221; . 1 12ABCS d A BC BC m = = s . Du = xy ra khi0 m =S:0 m =www.VNMATH.com11 V d 4) Cho hm s 4 22 2 y x mx = +c th (Cm). Tm tt c cc gi tr ca tham s m th (Cm) c 3 im cc tr to thnh 1 tam gic c ng trn ngoi tip i qua 3 9;5 5D| | |\ . Gii: C( )3' 4 4 0 0; 0 y x mx x x mm = = = = > . Vy cc im thuc ng trn (P) ngoi tip cc im cc tr l( )( ) ( )2 23 90; 2 , ; 2 , ; 2 , ;5 5A B m m C m m D| | + + |\ .. Gi( ) ; I xyl tm ng trn (P) ( ) ( )( )2 22 22 2 2 222 23 1 02 2 0; 1; 0( ), 12 2x y IA IDIB IC x y x m x y m L mIB IAx m y m x y + = = = = = = = = = + + + = + Vy1 m =l gi tr cn tm. Phn hai: Cc bi ton lin quan n tip tuyn v cc ng tim cn*) Xt hm s( ) y f x = .Gi s 0 0( ; ) Mx yl tip im khi tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1) ( Ch rng trong trng hp tng qut ta thng biu din 0ytheo dng 0( ) f x ) V d: Xt im M bt k thuc th hm s 2 11xyx=+ khi im M c to l 0002 1( ; )1xMxx+ *) Ta gi h s gc ca tip tuyn ti tip im M l 0'( ) k f x =*) ng thngA bt k c h s gc k i qua 0 0( ; ) Mx y c dng 0 0( ) y kx x y = + . iu kin A l tip tuyn ca hm s y=f(x) l h phng trnhsau c nghim 0 0( ) ( )'( )kx x y f xk f x + = = Khi snghim ca h cng chnh l s tip tuyn k c t im M n th hm s y=f(x) *) Mi bi tonvit phng trnh tip tuyn u quy v vic tm tip im sau vit phng trnhtheo (1) *) Cc dng cu hi thng gp trong phn ny l1) Vit phng trnh tip tuyn bit tip tuyn song song vi ng thng y=ax+b: + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1). Tip tuyn ti M c h s gc l 0'( ) k f x =+ Tip tuyn song song vi ng thng y=ax+b nn 0'( ) k f x a = = . Gii phng trnhtm 0xsau vit phng trnh tip tuyn theo (1) www.VNMATH.com12 Ch : iu kin cn tip tuyn ti A song song vi tip tuyn ti B l '( ) '( )A BA Bf x f xx x= = V d 1) Cho hm s 2 11xyx+=+. Vit phng trnh tip tuyn ca th (H) bit tip tuyn cch u hai im A(2;4), B(-4;-2) Gii : Gi 0xl honh tip im 0( 1) x = , PTTT l ( ) ( )0200 02 1 111xyxx x x+= +++ V tip tuyn cch u 2 im A,B nn tip tuyn i qua trung im I ca AB hoc song song vi AB hoc trng vi AB. Nu tip tuyn i qua trung im I(-1;1) ca AB th ta c:( )( )00 0 2002 11 1 111xx xxx= + =++ Suy ra phng trnh tip tuyn l 1 54 4y x = +Nu tip tuyn song song vi AB hoc trng vi AB th tip tuyn c h s gc l( )020002 ( 4) 11 12 4 ( 2)1xkxx= = = = = + Vi 00 x =ta c PTTT l1 y x = + ; vi 02 x = ta c PTTT l5 y x = +Vy c 3 PTTT tha mn 1 5; 1; 54 4y x y x y x = + = + = +V d 2) Cho hm s 12xyx =+ Tm trn th (C) 2 im A v B sao cho8 AB = , tip tuyn ca th (C) ti A v B song song vi nhau. Gii : Gi s im cn tm l 1 1; , ;2 2a bAa Bba b | | | | ||+ +\ . \ . theo gi thit ta c h: ( ) ( )( )( )( )22' '41 18 11 81 12 4a bf a f ba ba ba ba ba bab a b = =+ = | || | + = | + =+ +|\ .|+ + +\ . www.VNMATH.com13 ( )44116 4 1 8 14a ba bab abab+ = + = | | + = =|\ . t tm c A,B V d 3) Cho hm s 2(3 1)ym x m mx m=+ ++ (Cm) Tm m tip tuyn ti giao im ca (Cm) vi trc Ox song song vi ng thng (d): 1 y x = +Gii : Ta c 224'( )myx m=+ Giao im ca (Cm) v trc Ox l 2( ; 0)3 1m mAm+. Tip tuyn ti A ca (Cm) song song vi 2213 11 ' 1 113 1 25mm m my x ym m m= | | + | |

= + = = ||

+ = \ .\ . Khi m=1. Phng trnh tip tuyn l1 y x = +(loi) v tip tuyn trng vi ng thng (d) Khi 15m = . Phng trnh tip tuyn l :35y x = (TMK) KL : 15m = Qua v d ny cc em hc sinh cn lu : Kim tra iu kin khi tm ra gi tr tham s, y l sai lm hay mc phi ca hc sinh khi gii ton. V d 4) Cho hm s 33 2 y x x = +(C)Tm trn (C) cc im A,B phn bit sao cho cc tip tuyn vi (C) ti A,B c cng h s gc ng thi ng thng i qua A v B vung gc vi ng thng d: 5 0 x y + =Gii : Gi s cc tip tuyn vi (C) ti A,B c cng h s gc k. tn ti hai tip tuyn ti A,B phn bit th phng trnh 2' 3 3 y x k = =phi c hai nghim phn bit3 k > Ta cta cc im A,B tha mn h:( )2 3223 3 2 2 3 233 33 3xy x x y x xx kx k = + = + = = 2 22 2 2 2 2 23 3 33 3 3 3kx k ky x x y xx k x k | | | |= + = + = + || \ . \ . = = phng trnh ng thng AB: 2 23ky x| |= + |\ .. 2 1 93kAB d k = = (tha mn) www.VNMATH.com14 Vy ta cc im A,B tha mn: ( ) ( )3323 23 22; 4 , 2; 023 3 9y x xy x xA Bxx = += + = = V d 5) Cho hm s( ) ( )3 21 1 1 y x m x m x = + + +(1) Tm cc gi trca m th hm s ct Ox 3 im phn bit A(1;0), B, C sao cho cc tip tuyn ti B,C song song nhau. Gii: Xt phng trnh( )( )2 20 1 1 0( ) : 1 0 y x x mx gt pt x mx = = + =c 2 nghim phn bit khc 1 204 0mm= A = + >. Gi,B Cx xl nghim B Cx x =v B Cx x m + = . Yu cu bi ton( ) ( ) ' 'B Cy x y x =( ) ( ) ( ) ( ) ( )( )2 23 2 1 1 3 2 1 1 3 2 1 02 123B B C C B C B CB Cx m x m x m x m x x x x mmx x m m ( + + + + + + = + + = = = V d 6) Cho hm s( )2 2 11mx my Cx m += Cho A(1;2). Tm cc gi tr ca m sao cho tn ti ng thng qua A ct th Cm ti hai im phn bit M,N m cc tip tuyn ti M,N ca th song song vi nhau. Gii: Ta c: ( )23'1yx m= . Gi s( ) ( ) ( )1 1 2 2 1 2; , ;mMx y N x y C x x e = . Tip tuyn ti M v N song song ( ) ( )1 2 1 2 2 21 23 31 1 2 21 1x m x m x x mx m x m = = + = + (1) Ta thu c( )( ) ( )( )1 1 2 21 1 1 1 x x m x x m = v ch ( )1 2 1 2 1 21 ( 1) 1 1 2 x m x m x x x x = = + = . Cng vi (1)0 m = 2) Vit phng trnh tip tuyn bit tip tuyn vung gc vi ng thng y=ax+b + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1). Tip tuyn ti M c h s gc l 0'( ) k f x =+ Tip tuyn vung gcvi ng thng y=ax+b nn 01'( ) k f xa= = . Gii phng trnhtm 0xsau vit phng trnh tip tuyn theo (1) + Ch : iu kin cn tip tuyn ti A vung gc vi tip tuyn ti B l: '( ). '( ) 1A BA Bf x f xx x= = V d 1) Cho C(m): 3 2( ) 3 1 y f x x x mx = = + + +a)Tm m C(m) ct ng thng y=1 ti 3 im phn bit C(0;1), D, E. b)Tm m cc tip tuyn vi C(m) ti D v E vung gc vi nhau. www.VNMATH.com15 Gii: a) Xt( ) ( ) 1 Cm y = vi phng trnh tm honh giao im ( )3 2 2203 1 1 3 0 (0;1)( ) 3 0xx x mx x x x m Cgx x x m= + + + = + + =

= + + = Yu cu bi ton,D Ex x l 2 nghim phn bit khc 0 ca g(x)=0 99 4 090 4(0) 0 40m mmg mmA = > < = < = ==(*) b) o hm: 2( ) 3 6 y x x x m ' = + + .Vi iu kin 904m = v tip tuyn ca (Cm) ti mi im vung gc vi ng thng: 3 1 0 d x y + =Gii: Ta c h s gc ca: 3 1 0 d x y + =l 13dk = . Do 1 2, x xl nghim ca phng trnh y=-3 Hay( ) ( )2 22 2 1 3 2 3 2 2 1 3 1 x m x m x m x m + + = (1) Yu cu bi tonphng trnh (1) c hai nghim 1 2, x xtha mn 1 2. xx >0 ( ) ( )23 ' 1 2 3 1 013 11032m m mmm< A= + + > < < > Vy kt qu bi ton l3 m < v 113m < < . V d 3) Cho hm s 322 33xy x = + +(C) v ng thng (d) c h s gc k i qua A(0;3) Tm k ng thng (d) ct th (C) ti 3 im phn bit sao cho cc tip tuyn ti 3 giao im ct nhau to thnh mt tam gic vung. Gii: Honh giao im ca (C) v ng thng (d) l ( )32 22 3 3 6 3 03 3x xx kx x x k + + = + + =20( ) 6 3 0xgx x x k= = + =. iu kin l phng www.VNMATH.com16 trnh 2( ) 6 3 0 gx x x k = + =c 2 nghim phn bit khc 0. ' 0 9 3 0 3(0) 3 0 (0) 3 0 0k kg k g k kA> + > > = = = = = Ti x=0 tip tuyn song song vi trc Ox do 3 tip tuyn ct nhau to thnh mt tam gic vung th iu kin l 2( ) 6 3 0 gx x x k = + =c 2 nghim 1 2; x xsao cho 1 2'( ). '( ) 1 f x f x = ( ) ( )2 2 2 21 1 2 2 1 2 1 2 1 2 1 24 4 1 4 ( ) 16 1 0 x x x x x x x x x x x x + + = + + + + =Theo nh l Viets ta c 1 21 26. 3x xx x k+ = = Thay vo ta c: 2 24 159 72 48 1 0 9 24 1 03k k k k k k + + = + + = =Kt hp iu kin suy ra 4 153k = 3) Vit phng trnh tip tuyn bit tip tuyn i qua( ; )M MMx y+ Gi k l h s gc ca ng thngA i qua M . Phng trnh caA l( )M My kx x y = ++ iu kin A l tip tuyn ca y=f(x) l h sau c nghim ( ) ( )'( )M Mkx x y f xk f x + = =. Gii h tm x ta c honh ca cc tip im sau vit phng trnh tip tuynV d 1) Vit phng trnh tip tuyn i qua 19; 412A| | |\ . n( )3 2: ( ) 2 3 5 C y f x x x = = +Gii:ng thng i qua 19; 412A| | |\ . vi h s gc k c phng trnh 19412y k x| |= + |\ . tip xc vi( ) : ( ) C y f x = 19( ) 412( )f x k xf x k| |= + |\ . ' = c nghim ( )3 219 19( ) ( ) 4 2 3 5 6 1 412 12f x f x x x x x x x| | | |' = + + = + ||\ . \ . ( )( ) ( ) ( )( )( )( )( )( )( )21 1 12 2 23 3 319 171 2 1 6 1 1 4 1 012 2191 : 4 412192 : 4 12 15121 19 21 19: 4 48 12 32 12x x x x x x x xx t y y x yx t y y x y xx t y y x y x| | | | = + = ||\ . \ .

| |' = = + = | \ .

| |' = = + = |\ .

| | | |' = = + = +

||\ . \ . 4)Vit phng trnh tip tuyn bit tip tuyn to vi trc Ox mt gc www.VNMATH.com17 + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1). Tip tuyn ti M c h s gc l 0'( ) k f x =+ Tip tuyn to vi trc Ox mt gc 000'( ) tan'( ) tan'( ) tanf xf xf x= = = Gii tm 0xsau vit phng trnh tip tuyn theo (1). V d 1) Cho (C):3 21xyx=. Vit phng trnh tip tuyn ca (C) to vi trc honh gc 450 Gii: Do tip tuyn ca (C) to vi Ox gc 450 nn h s gc k ca tip tuyn tho mn 045 1 1 k tg k = = = . V ( )21( ) 0 11y x xx' = < = nn k=-1. honh tip im l nghim ca phng trnh ( )1 122 20 21( ) 1 12 41x yy xx yx= =' = = = = Phng trnh tip tuyn ti x1=0 l y=-1(x-0)+2=-x+2 Phng trnh tip tuyn ti x2=2 l y=-1(x-2)+4=-x+6. V d 2) Cho hm s 32( 1)xyx+=+ c th l (H).Vit phng trnh tip tuyn ti M trn (H) sao cho tip tuyn ct Ox, Oy ti A, B v ng trung trc ca AB i qua gc ta Gii: Do tam gic OAB vung ti O v trung trc ca AB i qua gc ta nn tam gic OAB vung cn ti O suy ra tip tuyn to vi Ox gc 450 Suy ra ( )0 0 0204'( ) 1 0 24 1f x x vxx= = = =+ T vit c 2 phng trnh tip tuyn l 32y x = + v 52y x = 5) Vit phng trnh tip tuyn bit tip tuyn to vi ng thng y=ax+b mt gc + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1). Tip tuyn ti M c h s gc l 0'( ) k f x =+ Tip tuyn to vi ng thng y=ax+b mt gc tan1tan1tan1k ak akak a kaka =

+ = +

=

+ (Vi 0'( ) k f x = ) Gii tm 0xsau vit phng trnh tip tuyn theo (1). V d 1) Cho (C):4 31xyx=. Vit phng trnh tip tuyn to vi( ) : A y=3x gc 450. Gii: Gi s tip tuyn c h s gc k, khi do tip tuyn to vi( ) A :y=3x gc 450 nn www.VNMATH.com18 023 1 3345 113 1 3 1 .32kk kktgk k k k= = +

= =

= + = * Vi k=-2, xt ng thng y=-2x+m tip xc (C) 4 321xx mx = + hay 4x-3=(-2x+m)(x-1) c nghim kp ( ) ( ) ( )2222 2 3 0 2 8 3 012 28 0 6 2 2x mx m m mm m m + + = A = = A = + = = * Vi k=12 xt ng thng 12y x m= +tip xc (C) 4 3 11 2xx mx = + hay 2(4x-3)=(-x+2m)(x-1) c nghim kp ( ) ( ) ( )222 7 2 6 0 2 7 4 2 6 0 x x x m m m + = A = =24 36 73 0 m m A = + =v nghim. Vy ch c 2 tip tuyn2 6 2 2 y x = + to vi y=3x gc 450. 6) Vit phng trnh tip tuyn bit tip tuyn ct hai trc to ti A, B sao cho tam gicOAB vung cn hoc tam gic OAB c din tch bng mt s cho trc. + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) y f x x x y = + (1). Tip tuyn ti M c h s gc l 0'( ) k f x =+ Tip tuyn ct 2 trc Ox, Oy ti A, B th tam gicOAB lun vung, OAB l tam gicvung cn th tip tuyn phi to vi Ox mtgc 045 =v tip tuyn khng i qua gc to + Vit phng trnh tip tuyn theo dng (4). Sau ch chn nhng tip tuyn khng i qua gc to + Nu yu cu l tip tuyn ct Ox, Oy to thnh tam gicc din tch cho trc th ta tm cc giao im A,B sau ta tnh din tch tam gic vung OAB theo cng thc 1.2OABS OAOBA=V d 1) Vit phng trnh tip tuyn ca th hm s 22xyx= bit tip tuyn ct , Ox Oyln lt ti A,B m tam gic OAB tha mn:2 AB OA = . Gii: Cch 1: Gi( )( )0 0 0; , Mx y x =thuc th hm s. PTTTd ti M c dng: ( )( )002002 422xy x xxx = . Do tip tuyn ct trc, Ox Oyti cc im A,B v tam gic OAB c2 AB OA =nn tam gic OAB vung cn ti O. Lc tip tuyn d vung gc vi 1 trong hai ng phn gicy x =hocy x = +TH1: d vung gc vi ng phn gicy x =c: ( )0 02041 0 42x xx= = v = www.VNMATH.com19 Vi 00 : x d y x = = (loi) Vi 04 : 8 x d y x = = + +TH2: : d vung gc vi ng phn gicy x = c: ( )20412 x= PT v nghim Vy c 1 tip tuyn tha mn yu cu bi ton: 8 d y x = +Cch 2: Nhn xt tam gic AOB vung ti O nn ta c:( )1sin sin4 2OAABOAB= = =nn tam gic AOB vung cn ti O. PTTT ca (C) ti( )0 0; M x y =c dng:( )( )002002 422xy x xxx= +. D dng tnh c 20;02xA| |=| |\ . v ( )202020;2xBx| | |= |\ . Yu cu bi ton lc ny tng ng vi vic tm0x l nghim ca phng trnh: ( )( )2023 00 02024 022xxx xx= = Vi 00 x =ta c PTTT l:0 y x + =Vi 04 x =th PTTT l:4 y x = +V d 2) Cho hm s 3 24 1(2 1) ( 2)3 3y x m x m x = + + + + (Cm) Tm m tip tuyn ti giao im ca (Cm) vi trc tung ct hai trc ta Ox, Oy ti A, B sao cho tam gic OAB c din tch bng 118 Ta c 1(0; )3Btip tuyn ti B ca (Cm) l 1( 2)3y m x = + +(d) . ng thng (d) ct trc Ox ti 1( ; 0)3 6Am+ Din tch tam gic OAB l 11 1 1 1 1. . . 2 13 2 2 3 3 6 18mS OAOB mm m= = = = + = = + 7) Vit phng trnh tip tuyn bit tip tuyn bit tip tuyn ct 2 ng tim cn to thnh mt tam gic c din tch cho trc hoc to thnh mt gc cho trc. + Xt hm s y=f(x). Gi 0 0( ; ) Mx yl tip im, suy ra tip tuyn ti M c dng 0 0 0'( )( ) ( ) y f x x x f x = + . + Tm cc giao im ca tip tuyn vi cc ng timcn sau cn c vo iu kin gii quyt + Nu yu cu l tip tuyn ct 2 tim cn ngang v tim cn ng ti A, B m tam gic IAB vung cn ( Vi I l giao im 2 tim cn) th ta quy v vic vit phng trnh tip tuyn bit tip tuyn to vi tim cn ngang mt gc 045 ) Ch rng tip tuyn khng c i qua giao im 2 ng tim cn v khi s khng hnh thnh mt tam gic) www.VNMATH.com20 + Nu yu cu l tip tuyn ct tim cn ng v tim cn ngang ti A, B to thnh tam gic IAB c din tch cho trc th ta tm cc giao im A, B sau dng cng thc 1.2OABS IA IBA=+ Ch : Gc to bi tip tuyn v ng tim ngang hoc tim cn ng cng chnh l gc to bi tip tuyn v cc trc Ox, Oy V d 1) Cho h s 2 3 mxyx m+=. Gi I l giao im ca hai tim cn. Tm m tip tuyn bt k ca hm s ct hai tim cn ti A,B sao cho din tch tam gic IAB bng64.Gii: D thy th hm s cho c ng tim cn ng l ng thngx m =v ng tim cn ngang l2 y m = . Ta giao im ca hai ng tim cn l: ( ) , 2 I m mGi 0002 3;mxM xx m| | + |\ . (vi 0x m = ) l im bt k thuc th hm s cho. PTTT ca th hm s ti im ny l ( )( )2002002 3 2 3 mx my x xx mx m+ += + Tip tuyn ny ct tim cn ng ti 2002 2 6;mx mAmx m| |+ + | |\ . v ct tim cn ngang ti ( )02 ; 2 B x m m . Ta c2 200 00 02 2 6 4 62 ; 2 2mx m mIA m IB x m m x mx m x m+ + += = = = Nn din tch tam gic IAB l 21. 4 62S IAIB m = = +Bi vy yu cu bi ton tng ng:2584 6 642m m + = = V d 2) Cho hm s.1xyx= Vit PTTT ca th (H) ca hm s cho bit tip tuyn to vi hai ng tim cn mt tam gic c chu vi bng ( )2 2 2 + . Gii: Cch 1: ng tim cn ca th l1, 1 x y = = . Gi PTTT ca (H) ti( )0 0; Mx yl: ( )( )0 0200111x x xyxx = + Khi 0 00 01 11 1;1 1x xx y Ax x| | + += = | \ .. Khi( ) ( )0 01 2 1 2 1;1 ; 1;1 y x x B x I = = www.VNMATH.com21 ( )( )( )( ) ( )( )( )( )( )( )( )( )( )220 00 00 02 40 0 00220 01 11 2 2 2 2 1 2 2 21 12 2 1 1 4 2 2 2 11 02 1 2 1 2 2 1 2 2 2 0ABCx xP IA IB AB x xx xx x xx Lx x| | + + = + + = + + + = + | \ . + + + = + = + + + + = Cch 2: Phng trnh tim cn ng1 x = , phng trnh tim cn ngang1 y =Gi;1aM aa| | |\ ., PTTT ti ( )( )21:11aM y x aaa= + Ta giao im ca tip tuyn v tim cn ng l 11;1aAa+ | | |\ . Ta giao im ca tip tuyn v tim cn ngang l( ) 2 1;1 B a Chu vi tam gic IAB l( )( )222 12 1 2 1 4 2 211C IA IB AB a aaa= + + = + + + > + Du = xy ra khi1 1 a =tc0; 2 a a = = . Vi0 a y x = = Vi2 4 a y x = = +KL:; 4 y xy x = = +l 2 tip tuyn cn tm. V d 3) Cho hm s( )3 21xy Cx=+. Gi I l giao ca 2 ng tim cn ca th. Vit PTTT d ca th hm s bit d ct tim cn ng v tim cn ngang ln lt ti A v B tha mn: 5cos26BAI =Gii: Xt im( ) ( )0 0 0; , 1 Mx y x C = el tip im ca tip tuyn d. PTTT ti d c dng: ( )( )002003 2 511xy x xxx = ++ Do tip tuyn d ct tim cn ng, tim cn ngang ln lt ti A v B vIAB Ac 5cos26BAI =nn 221 1 1 tan 1 tan tan 525 5 cosBAI BAI ABIBAI= = = =Li c tan ABIl h s gc ca tip tuyn d m( )( )0205' 02y xx= >+ nn ( )( )20 0 02055 1 1 0 21x x xx= + = = v = + Vi 00 x =c PTTT d: 5 2 y x = www.VNMATH.com22 Vi 02 x = c PTTT d: 5 2 y x = +Vy c 2 tip tuyn tha mn yu cu bi ton c pt nh trn. V d 4) Cho hm s : 2x 1yx 1=+ c th l( ) C . GiI l giao im ca hai ng tim cn ca( ) C .Tm trn th( ) CimM c honh dng sao cho tip tuyn tiM vi th( ) C ct hai ng tim cn tiA vBtho mn :2 240 IA IB + =Gii:TC( )1d: 1 x = ,TCN( )2: 2 d y = ( ) 1; 2 I .Gi0002 1;1xM xx| | |+\ .( )( )0, 0 C x e > Phng trnh tip tuyn vi( ) C ti( )( )( )00 2002 1 3: :11xM y x xxxA = +++ ( ) ( ) ( ) ( ) ( ) { }01 2 002 41; , 2 1; 21xd A d B xx | | A = A = + `|+ \ . ) ( )( )( ) ( )24 20 22 2 0 0000364 1 401 10 1 9 01 4000xx xx IA IBxx+ + = + + + = + + = > > 02 x =( )01 y = ( ) 2;1 M . 8) Vit phng trnh tip tuyn bit tip tuyn ct tim cn ng, tim cn ngang ti A, B m chu vi tam gic IAB nh nht *) gii quyt dng bi tp ny hc sinh cn nm c mt kt qu quan trng sau: (Tronghm s phn thc bc nht trn bc nht tip tuyn bt k ct 2 tim cn ti A,B th din tch tam gic IAB khng i). Vn dng kt qu ny ta c 2 22 . 2 . (2 2) .IABC IA IA AB IA IB IA IB IA IB IA IB IAIBA= + + = + + + > + = + . V din tch tam gic IAB khng i suy ra IA.IB khng i. T ta c Chu vi tam gic IAB min khi IA=IB. Gii iu kin tm M sau vit phng trnh tip tuyn V d 1) Cho hm s 21xyx=+. Vit PTTT ca th bit tip tuyn ct 2 tim cn ti A,B sao cho bn knh vng trn ni tip tam gic IAB ln nht. vi I l giao 2 tim cn. Gii: th hm s cho c tim cn ng l ng thng1 x = v tim cn ngang l ng thng1 y = . Giao im hai ng tim cn( ) 1;1 I . Gi s tip tuyn cn lp tip xc vi th ti im c honh 0x , PTTT c dang: ( )( )002002 311xy x xxx= +++ Tip tuyn ct tim cn ng1 x = ti im 0051;1xAx| | |+\ . v ct tim cn ng ti imwww.VNMATH.com23 ( )02 1;1 B x + . Ta c: ( )00 00 05 61 ; 2 1 1 2 11 1xIA IB x xx x= = = + = ++ + Nn 006. .2 1 121IAIB xx= + =+. Do vy din tch tam gic IAB l 1. 62S IA IB = =Gi p l na chu vi tam gic IAB, th bn knh ng trn ni tip tam gic ny l 6 Srp p= =Bi vy, r ln nht khi v ch khi p nh nht, mt khc tam gic IAB vung ti I nn: 2 22 2 2 . 4 3 2 6 p IA IB AB IA IB IA IB IA IB IA IB = + + = + + + > + + = +Du = xy ra khi( )201 3 1 3 IA IB x x = + = = Vi1 3 x = ta c tip tuyn ( ) 1: 2 1 3 d y x = + +Vi1 3 x = +ta c tip tuyn ( ) 2: 2 1 3 d y x = + V d 2) Cho Hypebol (C): 2 11xyx= v im M bt k thuc (C). Gi I l giao im ca tim cn. Tip tuyn ti M ct 2 tim cn ti A v B. a)CMR: M l trung im ca AB b)CMR: dt ( ) onst IAB c A =c)Tm M chu vi( ) IAB A nh nht. Gii:TC: x=1 TCN: y=2 Giaoim 2 tim cn l I(1;2) y =2 1 121 1xx x= + Gi M 1, 21mm| |+ e |\ .(c). Tip tuyn ti M l (t): y = ,y (m) (x-m) + y(m) 21 1( ) : ( ) 2( 1) 1t y x mm m = + + * (t) (TC: x =1) = A21, 21 m| |+ |\ .;(t) (TCN: y = 2) = B(2m 1, 2) Ta c :2A BMx xm x+= =v A,M,B thng hng nn M l trung im AB * dt( AIAB)=12IA . IB =12A I B Iy y x x 1 2 1 22( 1) .2( 1) 22 1 2 1m mm m= = = (vdt) Ta c IA . IB = 4 ;Chu vi ( AIAB) = IA + IB + AB=2 22 . 2 . 2(2 2) IA IB IA IB IA IB IAIB + + + > + = +www.VNMATH.com24 Du bng xy raIA = IB = 2 1 1 m = 120 (0, 1)2 (2, 3)m Mm M=

= 9) Tm iu kin qua im( ) ;M MMx ycho trc k c n tip tuyn n th y=f(x) + Xt ng thngA c h s gc k i qua im M( ) : PT A ( )M My kx x y = ++ iu kin A l tip tuyn ca y=f(x) l h sau c nghim ( ) ( )'( )M Mkx x y f xk f x + = =(*) + qua im M k c n tip tuyn n th th h (*) phi c n nghim th phng trnh (2) vo (1) dng phng php hm s tm iu kin+ Ch : Trong vic xc nh to M hc sinh cn linh hot VD: im M thuc ng thng y=2x+1 th M( ; 2 1) a a + , im M thuc ng thng y=2( ; 2) Ma V d 1) Cho th hm s (C):( )4 21 y f x x x = = + . Tm cc im A eOy k c 3 tip tuyn n th (C). Gii: Ly bt k A(0;a)e(C). ng thng i qua A(0;a) vi h s gc k c phng trnhy=kx+a tip xc vi th (C) ( )( )f x kx af x k= + ' = c nghim (*) iu kin cn: rng( ) ( ) ( ) f x f x x R f x = e l hm chn th (C)nhn Oy lm trc i xng. Do A(0;a)etrc i xng Oy nn nu t A(0;a) k c bao nhiu tip tuyn n nhnh bn tri ca (C) th cng k c by nhiu tip tuyn dn nhnh bn phi ca (C). Suy ra tng s cc tip tuyn c h s gc k= 0 lun l 1 s chn. Vy d t A(0;a)k c 3 tip tuyn dn (C) th iu kin cn l h phng trnh (*) c nghim k=0. Th k=0 vo h (*) 4 22 30; 11 11 3;4 2 02 4x ax x kxx ax x= = + = + = = = iu kin :Nu a=1 th (*)( )( )( )4 2 3 4 2332 2224 2 1 14 24 20; 00; 03 1 01 2;1 2; 3 3 32 13 31 2;3 3 3x x x x x x x kxx x kx x kx kx kx xx kxx kk x xx k = + = + = =

= =

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= ==

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Vy t A(0;1) k c 3 tip tuyn n (C) www.VNMATH.com25 Nu 34a =th (*) ( )( ) ( )4 2 4 2 33 34 2 422 23 31 1 4 24 44 2 4 21 113 04 222 1 2 1 0x x kx x x x x xx x k k x xx x xxk x x k x x k + = + + = + = = = == = = = Vy t 30;4A| | |\ . ch k c ng 1 tip tuyn n (C). Kt lun: T cc iu kin cn v p s: A(0;1) V d 2) Tm trn ng thng y=2x+1 cc im k c ng 1 tip tuyn n (C):31xyx+=. Gii: Ly bt k A(a;2a+1)ey=2x+1. ng thng i qua A(a;2a+1) vi h s gc k c phng trnh y=k(x-a)+2a+1 tip xc vi( ) ( )3 3: 2 11 1x xC y k x a ax x+ += + + = hay ( ) ( ) 2 1 1 3 kx ak a x x( = + c nghim kp ( ) ( )21 2 2 4 0 kx a k a x ak a(( + + + = c nghim kp 0 k = v ( ) ( )21 2 4 2 4 0 a k a k ak a(( A = + + = 0 k = v ( ) ( )22 2 2( ) 1 . 4 4 . 4 0 gk a k a a k a = + =Qua A(a;2a+1) k c ng 1 tip tuyn n (C) ( ) 0 gk = c ng 1 nghim kp k 0 =( )( )2 22 2032 2 0; (0) 4 0132 2 0; (0) 4 02111 0 16 4 04aa a g aaa a g aaaa k k=' A= > = == ' A= = = = == = + = = vy c 4 im ( ) ( ) ( ) ( )1 2 3 41; 1 , 0;1 , 1; 3 , 2; 5 A A A A nm trn dng thng y=2x+1 v k c ng 1 tip tuyn n th (C). V d 3) Cho hm s 3 22 ( 1) 2 y x x m x m = + +(Cm) Tm m t im M(1;2) k c ng hai tip tuyn n (Cm) Gii: Gi k l h s gc ca tip tuyn ta c phng trnh tip tuyn l(d) :( 1) 2 y kx = + . V (d) l tip tuyn nn h phng trnh sau c nghim 3 22( 1) 2 2 ( 1) 23 4 ( 1)y kx x x m x mk x x m = + = + += + 3 22 5 4 3( 1) 0 x x x m + = qua M k c ng hai tip tuyn n (Cm) th phng trnh 3 2( ) 2 5 4 3( 1) 0 f x x x x m = + =(*) c ng hai nghim phn bit. Ta c www.VNMATH.com26 21'( ) 6 10 4 '( ) 023xf x x x f xx=

= + = = . T tnh c hai im cc tr ca hm s l ( )2 1091; 4 3 , ; 33 27A m B m| | |\ .. Ta thy phng trnh (*) c ng hai nghim phn bit khi mt trong hai im cc tr nm trn trc honh. T tm c 43m =hoc 10981m = V d 4) Tm trn trc honh cc im k c 3 tip tuyn n th (C).33 2 y x x = + +Gii: Ly bt k A(a;0)e Ox. ng thng i qua A(a;0) vi h s gc k c phng trnhy=a(x-a) tip xc vi (C):y=f(x)H phng trnh( ) ( )( )f x k x af x k = ' = c nghim ( )( )( ) ( ) ( )3 22( ) ( )( ) ( ) 0 2 3ax 3 2 01 2 3 2 3 2 0 1 ( ) 0f x f x x af x f x x a x ax x a x a x gx' = ' = + + = ( + + + = + = T im A(a;0) k c 3 tip tuyn n (C)g(x)=0 c 2 nghim phn bit v khc (-1) ( )( )( )23 2 3 6 021 ( 1) 6 1 03aa aa g a> A = + > = < = + = Phn ba: Cc bi ton v s tng giao ca 2 th 1) Cc bi tp lin quan n php bin i th + T th y=f(x) suy ra th y=|f(x)| bng cch: Gi nguyn phn th ca y=f(x) nm trn trc Ox; Ly i xng ca phn th y=f(x) nm di trc Ox qua trc Ox. + T th y=f(x) suy ra th y=f(|x|) bng cch: Gi nguyn phn th y=f(x) nm bn phi trc Oy, Ly i xng ca phn th bn phi Oy qua trc Oy( Ch y=f(|x|) l hm chn nn nhn trc Oy lm trc i xng) + T th y=f(x) suy ra th y=|h(x)|.g(x) vi h(x).g(x)=f(x) bng cch. + Ta thy ( ) ( ) 0| ( ) | . ( )( ) ( ) 0f x khihxy hx gxf x khi x> = = < T ta suy ra cch v th hm s | ( ) | . ( ) y hx gx =nh sau:Ly phn th y=f(x) khi( ) 0 hx > . Ly ixng qua trc Ox phn th y=f(x) khi( ) 0 hx A ss hoc '( ) 1 21 20 '( ) 0. 0 ( ). ( ) 0f xCD CTf x x x x xf f f x f xA > = = v = > >

+ Hm s : y=ax3+bx2+cx+d ct trc Ox ti 2 im phn bit khi v ch khi f(x) c 2 nghim phn bit 1 2; x xv1 2( ) ( ) 0 f x f x =+ Hm s : y=ax3+bx2+cx+d ct trc Ox ti 3 im phn bit khi hm s c cc i, cc tiu v gi tr cc i, cc tiu tri du nhau'( ) 0 f x =c 2 nghim phn bit 1 2; x x v 1 2( ). ( ) 0 f x f x + > > ng vi mi m. Khi 211' 01x myx m= + = = Ta c: ( ) ( )( )( )( )( )( )( )( )( )222212 2 21 21 2 11 1' . 2 1 13 31 3. 1 3 2 1y m m my f x x m x m my m my y m m m m= + | |= + |\ .= = ( ) ( )( )( )2 21 D2 20 1 0 1 00 13 2 1 0(*)Cf m mx x mm m m= < > > > < Lp bng xt du (*) kt hp iu kin1 m >Suy ra tp hp gi tr m tha mn l3 1 2 m < < +www.VNMATH.com28 V d 2) Chng minh rng phng trnh 3 2 2 33( 1) 3( 1) 1 0 x m x m x m + + + + + + =lun c nghim duy nht. Gii ; Xem phng trnh 3 2 2 33( 1) 3( 1) 1 0 x m x m x m + + + + + + =l phng trnh honh giao im ca 3 2 2 33( 1) 3( 1) 1 y x m x m x m = + + + + + +v trc honh. Ta c 3 21 1'. 23 3my y x mx m m+ | |= + + |\ . suy ra ng thng qua hai cc tr l 3 22 y mx m m = + phng trnh c nghim duy nht th th hm s 3 2 2 33( 1) 3( 1) 1 y x m x m x m = + + + + + +ct trc Ox ti mt im duy nht.Tc l ( ) ( )3 2 3 218 8 0' 018 8 0' 0 (**). 0 2 2 0CD CT CD CTmmy y mx m m x m m s As

> A>

> + + > Theo nh l viet: 22( 1). 1CD CTCD CTx x mx x m+ = + = + Thay vo (**) ta c 2 2 2 3229922994 ( 1) ( 1) (4 1) 0mmmmmm m m m m

s s > > + + + > Vy vi mi m phng trnh lun c nghim duy nht. V d 3) Gi s th hm s 3 26 9 y x x x d = + +ct trc Ox ti 3 im phn bit 1 2 3x x x < < . Chng minh 1 2 30 1 3 4 x x x < < < < < > < Gi( ) ( )1 1 2 2; 1 , ; 1 Bx x Cx x + + . B v C i xng vi nhau qua ng phn gic th nht th:1 2 1 21 21 2 2 113 21 11 2 3x y x xx x m my x x x= = + + = = = = = + So snh vi K, thy khng tm c m tha mn. V d 6) Cho hm s 3 23 4 y x x = + Gi kdl ng thng i qua im( ) 1; 0 Avi h s gck ( ) k e . Tmk ng thng kdct th (C) ti ba im phn bit v hai giao im, B C ( BvCkhcA) cng vi gc to O to thnh mt tam gic c din tch bng 1. Gii: :kd y kx k = +(hay0 kx y k + = ). Pt honh giao im ca kdv (C):( ) ( )23 23 4 1 2 0 1 x x kx k x x k x ( + = + + = = hoc( )22 x k =kdct (C) ti 3 im phn bit 09kk> = (d) ct (C)ti ( )( ) ( )1;0 , 2 ;3 , 2 ;3 A B k k k k C k k k k + + . ( ) ( )222 1 , , ,1kkBC k k d OBC d O dk= + = =+2 321. .2 . 1 1 1 1 121OBCkS k k k k k kkA= + = = = =+ V d 7) Cho hm s 3 22 3( 1) 2 y x mx m x = + + +(1),m l tham s thc Tmm th hm s ct ng thng: 2 y x A = +ti 3 im phn bit(0; 2) A ; B; C sao cho tam gicMBC c din tch2 2 , vi(3;1). MGii: www.VNMATH.com31 Phng trnh honh giao im ca th vi( ) A l:3 22 3( 1) 2 2 x mx m x x + + + = +20 2( ) 2 3 2 0(2)x ygx x mx m= = = + + = ng thng( ) A ct th hm s (1) ti ba im phn bit A(0;2), B, C Phng trnh (2) c hai nghim phn bit khc 0 22' 0 3 2 0 1(0) 0 3 2 023mm m mg mm > A> + > < = = = Gi( )1 1; Bx y v( )2 2; Cx y , trong 1 2, x xl nghim ca (2); 1 12 y x = +v 1 22 y x = +Ta c( )3 1 2; ( )2h d M+ = A =2 2.2 242MBCSBCh = = =M 2 2 2 22 1 2 1 2 1 1 2( ) ( ) 2 ( ) 4 BC x x y y x x x x( = + = + =28( 3 2) m m +Suy ra 28( 3 2) m m + =16 0 m = (tho mn) hoc3 m = (tho mn) 4) iu kin hm s bc 3 c 3 nghim lp thnh cp s cng Xt phng trnh 3 2ax 0 bx cx d + + + = . Gi s phng trnh c 3 nghim lp thnh cp s cng l 1 2 3; ; x x xkhi : 3 2 3 21 2 3 1 2 3 1 2 2 3 3 1 1 2 3ax ( )( )( ) [x ( ) ( ) ] bx cx d ax x x x x x a x x x x x x x x x x x x x + + + = = + + + + + V 3 nghim lp thnh cp s cng nn 1 3 2 223bx x x xa+ = = l nghim. Th vo phng trnh ta suy ra iu kin cn tm. V d 1) Cho( ) ( ) ( )3 2 2: 3 2 4 9 Cm y f x x mx mm x m m = = + + . Tm m C(m) ct Ox ti 3 im phn bit lp thnh cp s cng. Gii:iu kin cn: Gi s (Cm) ct Ox ti 3im phn bit l 1 2 3, , x x x .Khi :( )3 2 23 2 4 9 0 x mx mm x m m + + =c 3 nghim phn bit 1 2 3, , x x x( ) ( )( )( )3 2 21 2 33 2 4 9 x mx mm x m m x x x x x x x + + = ( ) ( ) ( )3 2 2 3 21 2 3 1 2 2 3 3 1 1 2 33 2 4 9 x mx mm x m m x x x x x x x x x x x x x x x x + + = + + + + + Suy ra( )1 2 3 1 3 2 2 23 3 m x x x x x x x x m = + + = + + = =Th 2x m =vo 2( ) 0 0 0 f x m m m = = =hoc1 m =iu kin : Vi m=0 th 31 2 3( ) 0 0 f x x x x x = = = = = (loi) Vi m=1 th 3 2( ) 3 6 8 0 f x x x x = + = ( )( )21 2 31 2 8 0 2; 1; 4 x x x x x x = = = =Kt lun: p s m=1. 5) iu kin hm bc 3 c 3 nghim lp thnh cp s nhn www.VNMATH.com32 Xt phng trnh 3 2ax 0 bx cx d + + + = . Gi s phng trnh c 3 nghim lp thnh cp s cng l 1 2 3; ; x x xkhi : 3 2 3 21 2 3 1 2 3 1 2 2 3 3 1 1 2 3ax ( )( )( ) [x ( ) ( ) ] bx cx d ax x x x x x a x x x x x x x x x x x x x + + + = = + + + + + V 3 nghim lp thnh cp s nhn nn 2 331 3 2 1 2 3 2 2axdx x x d x x x xa= = = = thay vo phng trnh ta suy ra iu kin cn tm V d 1) Cho( ) ( ) ( ) ( )3 2: 3 1 5 4 8. Cm y f x x m x m x = = + + + Tm m (Cm) ct Ox ti 3 im phn bit lp thnh 1 cp s nhn. Gii:iu kin cn: Gi s (Cm) ct Ox ti 3 im phn bit 1 2 3, , x x x Khi :( ) ( )3 23 1 5 4 8 0 x x x x x + + + =c 3 nghim phn bit 1 2 3, , x x x( ) ( ) ( )( )( )3 21 2 33 1 5 4 8 x x x m x x x x x x x x + + + = ( ) ( ) ( ) ( )3 2 3 21 2 3 1 2 2 3 3 1 1 2 33 1 5 4 8 x x x m x x x x x x x x x x x x x x x x + + + = + + + + + Suy ra: 31 2 3 2 28 2 x x x x x = = =Th 22 x =vo( ) 0 4 2 0 2 f x m m = = =iu kin : Vi m=2 th ( ) ( )( )( )3 21 2 37 14 8 0 1 2 4 0 1; 2; 4 f x x x x x x x x x x = + = = = = =Kt lun: p s m=2. 6) iu kin hm s bc bn c 4 nghim lp thnh cp s cng Xt phng trnh 4 2ax 0 bx c + + =(1) t 2( 0) t x t = > phng trnh (1)c 4 nghim lp thnh cp s cng th phng trnh 20 at bt c + + = (2) phi c 2 nghim dng phn bit 1 2, t t . Gi s (1 2) t t < khi 4 nghim ca (1) l 2 1 1 2, , , t t t t v 4 nghim lp thnh cp s cng nn ( ) 2 1 1 1 2 19 t t t t t t = = . p dng nh l vit cho phng trnh (2) ta c 1 21 21 29bt tact tat t+ = == Gii iu kin theo h phng trnh. V d 1) Cho( ) ( )4 2: 2 1 2 1. Cm y x m x m = + + +Tm m (Cm) ct Ox ti 4 im phn bit lp thnh 1 cp s cng. Gii: Xt phng trnh:( )4 22 1 2 1 0(1) x m x m + + + =t( ) ( )2 2; 2 1 2 1 0(2) t x f t t m t m = = + + =Yu cu bi ton( ) 0 f t =c 2 nghim 2 10 t t > >sao cho (1) c s nghim www.VNMATH.com33 Ta c 4 3 3 2 2 1 4 3 3 2x x x x x x x x x x = = = ( ) 2 1 1 1 2 1 2 13 9 0 t t t t t t t t = = = >Yu cu bi ton ( )22 12 11 2 2 12211 21110, 9 0 229. 2 1 0 99 2 12 1 019 2 15 15mmm t tt tt t m t tt mt t mmmt m> > ' A= > = > = = + > = = ++ = + >+ | | = + |= + \ .

2 1214 2949 9 32 16 0mmt tmm m> =

=

= = V d 2) (Bi ton tng giao hm bc 4) Tm m sao cho th hm s 4 24 y x x m = +ct trc honh ti 4 i phn bit sao cho din tchhnh phng gii hn bi (C) trc honh c phn trn bng phn di Gii: Phng trnh honh giao im ca th (C) v 4 2: 4 0 Ox x x m + = (1) t 20 t x = > . Lc c PT:24 0 t t m + = (2) (C) ctOxti 4 im phn bit khi pt (1) c 4 nghim phn bit(2) c 2 nghim phn bit()' 4 00 4 0 40mt S m iP mA= > > = >= < Gi( )1 2 1 2, 0 t t t t < | | = |\ . ng thng ct th (H) ti 2 im phn bit A, B khi ta c1 1 2 2( ; ); ( ; ) A x mx n Bx mx n + +vi x1; x2 l hai nghim ca g(x)=0 V d 1) Cho hm s 32xyx+= c th (H). Tm m ng thng: 1 d y x m = + +ti hai im phn bit A,B sao cho AOB nhn. Gii:Giao ca (H) v d c honh l nghim ca PT ( )231 2 2 5 02xx m x m x mx+= + + + + + = pt trn c 2 nghim phn bit th0; 2 x A > =( )224 16 0?2 2 2 2 5 0m mmm m + > = + + + = Gi( ) ( )1 1 2 2; 1 , ; 1 Ax x m Bx x m + + + +l hai giao im ca (H) v d. AOB nhn th: ( ) ( ) ( )( )( ) ( )2 2 22 2 22 1 1 221 2 1 22 1 12 1 1 0 3AB OA AB x x x m x mx x m x x m m< + < + + + + + + + + + < > Kt hp vi k ban u ta suy ra c gi tr ca m. www.VNMATH.com35 V d 2) Cho hm s 21x mymx=+ (1). Chng minh vi mi0 m = th hm s (1) ct ( ) : 2 2 d y x m = ti 2 im phn bit A,B thuc 1 ng (Hipebol) c nh. ng thng (d) ct cc trc, Ox Oyln lt ti cc im M,N. Tm m 3OAB OMNS S =Gii: Phng trnh honh ca giao im ca th hm s (1) v ng thng (d): 2 22 12 2 2 2 0,1x mx m mx mx m xmx m | |= = = |+\ .(2) Do0 m =nn (2) ( )212 2 1 0, f x x mx xm| | = = = |\ .(*) tn ti 2 im A,B th pt (*) phi c 2 nghim phn bit: 22' 2 01, 01 21 0A Bmx x mm fm mA= + >= =| | = + = | \ . Mt khc c 1.2A Bx x =nn A,B lun thuc mt ng (Hipebol) c nh K ( ) ,25O dmOH AB OH d = = . Li c2 2 ; 2 2A A B BAB d y x my x m e = = Theo Viet ta c: 12A BA Bx x mx x+ = = C ( ) ( ) ( ) ( )2 2 2 225 5 20 5 10A B A B A B A B A BAB x x y y x x x x x x AB m = + = = + = +V M,N l giao im ca d vi, Ox Oynn( ) ( ) ; 0 , 0; 2 Mm N mTheo gi thit 223 . 3 . . 5 10 35OAB OMN M NmS S OH AB OM ON m x y= = + =2 2 2 221. 5 10 3 2 2 3 2 92 5mm m m m m m m m + = + = + = = Vy vi 12m = l cc gi tr cn tm. V d 3) Tm trn (H):12xyx += cc im A,B sao cho di on thng AB bng 4 v ng thng AB vung gc vi ng thngy x =Gii: Do: : AB d y x ptAB y x m = = +Phng trnh honh giao im ca (H) v ng thng AB: ( ) ( ) ( )213 2 1 0 22xx m g x x m x m xx += + = + + + = =(1) tn ti 2 im A,B th pt (1) cn c 2 nghim phn bit,A Bx xv khc 2 www.VNMATH.com36 ( )( )( ) ( )( )( )22203 4 2 1 01 4 0;2 04 3 2 2 1 0g xm mm mgm mA > + + > + > = + + + = Theo Viet ta c: 32 1A BA Bx x mx x m+ = + = +. Li c;A A B By x y x m = = +M: ( ) ( ) ( )( ) ( ) ( )2 2 222 224 16 16 84 8 3 4 2 1 0 2 3 0 1 3B A B A B AB A A BAB AB x x y y x xx x xx m m m m m m= = + = = + = + + = = = v = + Vi3 m =thay vo pt (1) c:26 7 0 3 2 2 x x x y + = = = . Lc ny ta 2 im A,B l: ( ) ( )3 2; 2 , 3 2; 2 A B + hoc ( ) ( )3 2; 2 , 3 2; 2 B A + . + Vi1 m = thay vo pt (1) c:22 1 0 1 2 2 2 x x x y = = = . Lc ny ta 2 im A,B l ( ) ( )1 2; 2 2 , 1 2; 2 2 A B + + hoc( ) ( )1 2; 2 2 , 1 2; 2 2 B A + + Vy A,B l cc im nh trn tha mn yu cu bi ton. V d 4) Cho hm s 32xyx+=+ c th l (H). Tm m ng thng d: 2 3 y x m = +ct (H) ti hai im phn bit sao cho. 4 OAOB = vi O l gc ta Gii: Xt pt:( )232 3 2 3 1 6 3 02xx m x mx mx+= + + + + =+ (1) c 2 nghim phn bit khc (-2) Khi 29 30 33 0 m m A = + >iu ny xy ra vi mi m. Gi 2 nghim ca pt (1) l 1 2, x xth( ) ( )1 1 2 2; 2 3 , ; 2 3 Ax x m Bx x m + +C( )( )1 2 1 212 15 7. 4 . 2 3 2 3 4 42 12mOAOB x x x m x m m= + + + = = = V d 5) Cho hm s 2 11xyx=+ c th (C). Tm m ng thng d: y x m = +ct (C) ti hai im phn bit A,B, sao cho2 2 AB =Gii: Phng trnh honh giao im ca (C) v ng thng d: ( ) ( )22 11 1 01xx m f x x m x mx= + = + + + =+ (1) ( ) 1 x = d ct (C) ti 2 im phn bit A,B th pt(2) c 2 nghim phn bit, 1A Bx x = ( ) ( )( )21 4 1 01 1 1 1 0m mf m mA = + > = + + + = (*) Theo Viet ta c:( )221 1; , ; 1 4( 1) 41 7A BA A B BA Bx x m mA B d y x my x m AB m mx x m m+ = = e = + = + = + =

= + = www.VNMATH.com37 V d 6) Gi D l ng thng i qua A(1;0) v c h s gc k. Tm k D ct th 21xyx+= ti hai im phn bit M,N thuc hai nhnh khc nhau ca th v AM=2AN Gii: Do D l ng thng i qua A(1;0) v c h s gc k nn pt D:( ) 1 y k x = Phng trnh honh giao im ca D v th hm s cho l:( ) ( ) ( )221 2 1 2 0 11xk x kx k x xx+= + = = (1) t1 1 t x x t = = + . Lc pt (1) thnh: ( ) ( )( )221 2 1 1 2 0 3 0 k t k t k kt t + + + + = = (2) D ct th hm s cho ti hai im M,N thuc hai nhnh khc nhau ca th th pt(1) phi c 2 nghim 1 2, x xtha 1 21 (2) x x pt < < c 2 nghim 1 2, t ttha 1 20 3 0 0(*) t t k k < < < >V im A lun nm trong on MN v 1 22 2 2 3 AM AN AM AN x x = = + = (3) Theo Viet ta c: ( )( )1 21 22 1425kx xkkx xk+ + ==. T (3) v (4) 2 11 2;k kx xk k + = =Thay 1 2, x xvo pt (5) c: ( ) ( )22 12 23 2 03k kkk kk k+ = = =i chiu K (*) c 23k =l gi tr cn tm. Phn bn: Cc bi ton v khong cch gia quyt tt cc dng bi tp trong phn ny hc sinh cn nm chc cc vn sau: *) Khong cch gia hai im( ; ); ( ; )M M N NMx y Nx yl( ) ( )2 2N M N MMN x x y y = + *) Khong cch t im 0 0( ; ) Mx yn ng thng: ax+by+c=0 Al 0 0/2 2axMby cda bA+ +=+ Cc trng hp c bit: + NuA l ng thng x=a th / 0 Md x aA = + NuA l ng thng y=b th / 0 Md y bA = + Tng khong cch t M n hai trc to Ox, Oy l d=0 0x y +*) Khong cch gia ng thng v ng congCho ng thngA v ng cong ( C) . Ly im M bt k thuc ng cong ( C) v im N thuc ng thngA Khi ( /( ))minCd MNA= . T ta c cch tnh khong cch t ng thng: ax+by+c=0 An ng cong ( C) y=f(x) nh sau: www.VNMATH.com38 + Cch 1: Ly im M( )0 0; x ybt k thuc ( C) 0 0( ; ( )) Mx f x . Ta c 0 0/2 2axMby cda bA+ +=+ Sau tm min d theo x0 + Cch 2: Vit phng trnh tip tuyn t ca ng cong ( C) v tip tuyn song song viA. Sau tm tip im M( )0 0; x yca tip tuyn v ng cong. Khi khong cch gia ng thngA v ng cong ( C) cng bng khong cch gia M v ng thngA l 0 0/2 2axMby cda bA+ +=+ V d 1)Cho th ( )2 1:1xC yx+= v im A(-2;5). Xc nh ng thng (D) ct (C) ti 2 im B, C sao choAABC u. Gii: 2 11xyx+=: 1: 2TCDxTCN y= = phn gic ca gc to bi 2 tim cn (1): 3 y x = +( )2301yx' = < hm s nghch bin th (C) c dng nh hnh v. Do A(-2;5)(1) : 3 y x e = +l trc i xng ca (C) nn ng thng (D) cn tm phi vung gc vi (1) v (D) c phng trnh: y=x+m. Xt phng trnh: ( ) ( ) ( )22 13 1 01xx m g x x m x mx+= + = + + = Ta c( ) ( ) ( )2 23 4 1 1 12 0 g m m m A = + + = + > nn (D) lun ct (C) ti B, C phn bit v do tnh i xng A ABC cn ti A. Gi s( ) (1) D223 3 7; 22 2 2m m mI I AI + | | | |= = ||\ . \ .

Gi ( )( )( ) ( )2 2 1 11 1 21 2 1 2 1 22 22 2,2 2 4,Bx yy x mBC x x x x x xy x mCx y= + ( = = + = + ( ) ( ) ( )22 22 3 4 1 2 2 13 BC m m m m (= + + = + Ta cAABC u( ) ( )22 2 243 2 13 73BC AI m m m = + = ( )( )122: 114 5 05: 5D y xmm mmD y x = += + =

= = V d 2) Cho( )3 5:2xH yx=. Tm Me(H) tng khong cch t M n 2 tim cn ca (H) l nh nht. Gii:Ta c TC: x=2 TCN: y=3 www.VNMATH.com39 3 5 132 2xyx x= = + Ly 1;32M m Hm| |+ e |\ . Tng khong cch t M n cc ng tim cn l 12 3 2 22M M Md x y mm= + = + > ( Theo bt ng thc Cauchy) 1 (1; 2)1min 2 23 (3; 4) 2m Md mm M m= = = = V d 3) Tm trn mi nhnh ca th ( )4 9:3xC yx=cc im cch M1,M2 di 1 2, M Ml nh nht. Gii: ( ) : 3 4 3 34 9 34: 4 3 3 3TCDx xxyTCN y x x x= + = = = + = Gi ( )( )1 1 12 2 2,,M x yM x y(M1 thuc nhnh tri ca (C); M2 thuc nhnh phi ca (C)) t 1122334334, 0xyxy = = = + = > ( ) ( )2 221 2 2 1 2 1M M x x y y = + =( ) ( )( )( )2 22 2293 3 + | |+ + + = + + |\ . ( )( )( )( )cos222 29 9 91 2 1 4i ((| |= + + > + = +(( |+ +(( \ . M cos9 94 4 2 . 24i (| |+ > = ( |\ . 1 20min 24 2 6 3 9 M M = > = = = == to ( ) ( ) 1 23 3; 4 3 , 3 3; 4 3 M M + + V d 4) Tm im M trn th hm s 2 1( )1xy Hx+=+ sao cho khong cch t M n ng thng ( A): 4 8 0 x y + =l nh nht Gii:Ta c ( )21'1yx=+. Xt ng thng d l tip tuyn ca (H) vd song song vi ( A) www.VNMATH.com40 T vit c 2 phng trnh tip tuyn l 15:4 4xd y = +v 213:4 4xd y = +Hai tip im tng ng l 1 23 51; ; 3;2 2M M| | | | ||\ . \ . D dng tnh c( ) ( )1 2/ / d M d M A < A1M l im cn tmCh : Ngoi cch lp lun nh trn ta c th gii bi ton theo cch khc gi s 2 1( ; )1xMxx++ Sau tnh khong cch t M nA. V tm min theo phng php hm s vi bin x Cho hm s 1xyx= v im A(-1;1) V d 5) Tm m ng thng 1 y mx m = ct (C) ti hai im phn bit M,N sao cho 2 2AM AN +t gi tr nh nht. Gii: Xt phng trnh tng giao: 22 1 0 mx mx m + + = . ct ti hai im th phng trnh phi c 2 nghim phn bit khc 1.( )22 1 00' 1 0m m mmm mm m + + = thy trung im MN l I v I(1;-1) c nh. S dng chn im ta c:2 2 2 2 22 AM AN AI IM IN + = + +(do0 IM IN + = ) Ta c IA c nh, IM=IN. Ta thy biu thc min khi v ch khi MN min Tnh MN: ( ) ( ) ( ) ( )2 2 22 21 2 1 2 1 241 4 . 1 4 NM x x m x x x x m mm (= + = + + = Do m . 244 8 MN tt= + > . Vy m=-1 V d 6) Tm m ng thng y=mx-m+2 ct (C) 21xyx= ti 2 im phn bit A,B sao cho di AB nh nht. Gii:ng thng y=mx-m+2 ct (C) ti 2 im phn bit khi phng trnh 221xmx mx= + c 2 nghim phn bit khc 1. 2( ) 2 2 0 gx mx mx m = + =c 2 nghim phn bit khc 1 00(1) 0mg= A >=0 m > . Ta c 1 1 2 2( ; 2); ( ; 2 A x mx m Bx mx m + + ) ( )2 2 22 1 2 1 2 1; ( ) ( ) (1 ) AB x x mx x AB x x m = = + ( )( )22 21 2 1 24 ( 1) AB x x x x m = + +V x1;x2 l 2 nghim ca g(x)=0 nn ta c 1 2 1 222;mx x x xm+ = =218( ) 16 min 4 1 AB m AB mm = + > = =www.VNMATH.com41 Phn nm: Cc bi tp v KSHS CC DNG BI TP V KHO ST HM S Bin son GV Nguyn Trung Kin 0988844088 Phn mt: CC BI TP LIN QUAN IM CC I V CC TIU HM S Cu 1) Cho hm s1312 3+ + = m x mx x ya)Kho st v v th hm s khi m=1 b)Tm m hm s c cc i cc tiu v khong cch gia im cc i v cc tiu l nh nht Cu 2) Cho hm s1312 3 + = mx mx x ya)Kho st v v th hm s khi m= 1 b)Tm m hm st cc tr ti 2 1; x xtho mn82 1> x x Cu 3) Cho hm s3 72 3+ + + = x mx x ya)Kho st v v th hm s khi m= -8 b)Tm m hm s c ng thng i qua im cc i cc tiu vung gc vi ng thng y=3x-7 Cu 4) Cho hm s) 1 ( ) 2 3 2 ( ) 1 ( 32 2 3 + + = m m x m m x m x ya)Kho st v v th hm s khi m= 1 b)Tm m hm sc cc i cc tiu v ng thng i qua cc i cc tiu to vi ng thng541+= x ymt gc 450 Cu 5) Cho hm sm x m x x y + + =2 2 33a)Kho st v v th hm s khi m= 0 b)Tm m hm sc cc i cc tiu i xng qua ng thng 2521 = x y Cu 6) Cho hm s1 3 ) 1 ( 3 32 2 2 3 + + = m x m x x ya)Kho st v v th hm s khi m= 1 b)Tm m hm sc cc i cc tiu cch u gc to O. Cu 7) Cho hm s1 22 2 4+ = x m x ya)Kho st v v th hm s khi m= 1 b)Tm m hm s c 3 im cc tr l 3 nh ca tam gic vung cn Cu 8) Cho hm s1 12 9 22 2 3+ + + = x m mx x ywww.VNMATH.com42 a)Kho st v v th hm s khi m= 1 b)Tm m hm sc cc i cc tiu ng thi CTCD x x =2 Cu 9) Cho hm s 4 2 42 2 m m mx x y + + =a)Kho st v v th hm s khi m= 1 b)Tm m hm s c cc i cc tiu lp thnh mt tam gic u Phn hai: CC BI TON LIN QUAN N TIP TUYN V NG TIM CN Cu 1) Cho hm s13+ = m mx x y (Cm)a)Kho st v v th hm s khi m= 3 b)Tm m tip tuyn ti giao im cu (Cm) vi trc Oy chn trn hai trc to mt tam gic c din tch bng 8 Cu 2) Cho hm s1 32 3+ + + = mx x x y(Cm) a)Kho st v v th hm s khi m= 0 b)Tm m ng thng y=1 ct (Cm) ti 3 im phn bit C(0;1), D,E v cc tip tuyn ti D v E ca (Cm) vung gc vi nhau. Cu 3) Cho hm sx x y 33 =(C ) v ng thng y=m(x+1)+2 (d) a)Kho st v v th hm s (C) b)Chng minh rng ng thng (d) lun ct (C ) ti mtim c nh A. Tm m ng thng (d) ct (C ) ti 3 im A,M,N m tip tuyn ti M v N vung gc vi nhau\ Cu 4) Cho hm s) (12 3Hxxy=a)Kho st v v th hm s (H) b)Vit phng trnh tip tuyn ca (H) bit tip tuyn to vi Ox gc 450 c)Vit phng trnh tip tuyn ca (H) bit tip tuyn to vi 2 trc to mt tam gic cn d)Gi I l giao im 2 ng tim cn. Tip tuyn ti M bt k thuc (H) ct 2 tim cn ti A,B. Chng minh M l trung im AB e)Chng minh din tch tam gic IAB khng i f)Tm v tr M chu vi tam gic IAB nh nht Cu 5) Cho hm s) (2Hmxm xy+=a)Kho st v v th hm s khi m= 3 b)Tm m t A(1;2) k c 2 tip tuyn AB,AC n (Hm) sao cho ABC l tam gic u (A,B l cc tip im) Cu 6) Cho hm s) (3 2Hmm xmxy +=1)Kho st v v th hm s khi m=1 2)Tm m tip tuyn bt k ca hm s (Hm) ct 2 ng tim cn to thnh mt tam gicc din tch bng 8 www.VNMATH.com43 Cu 7) Cho hm s) (11 2Hxxy+=a)Kho st v v th hm s(H) b)Vit phng trnhng thng ct (H) ti B, C sao cho B, C cng vi im) 5 ; 2 ( A to thnh tam gic u Cu 8) Cho hm s) (12Hxxy+=a)Kho st v v th hm s cho b)Tm M thuc (H) sao cho tip tuyn ti M ca (H) ct 2 trc Ox, Oy ti A, B sao cho tam gic OAB c din tch bng 41 Cu 9) Cho hm s) (11 2Hxxy=a)Kho st v v th hm sb)Gi I l giao im 2 ng tim cn ca (H). Tm M thuc (H) sao cho tip tuyn ca (H) ti M vung gc vi ng thng IM. Cu 10) Cho hm s) (22Hxxy+=a)Kho st v v th hm s(H) b)Vit phng trnhtip tuyn ca (H) bit khong cch t tm i xng ca th hm s (H) n tip tuyn l ln nht. Cu 11) Cho hm s) ( 1 2 32 3C x x x y + + =a)Kho st v v th hm s b)Tm hai im A,B thuc th sao cho tip tuyn ca (C ) ti A, B song song vi nhau v di AB nh nht Cu 12) Vit cc phng trnhtip tuyn k t im|.|

\|4 ;1219An th hm s 5 3 22 3+ = x x y Cu 13) Tm im M thuc th hm s2 32 3 + = x x ym qua ch k c mt tip tuyn n th Cu 14) Tm nhng im thuc ng thng y=2 m t c th k c 3 tip tuyn n th hs 2 33x x y = Cu 15) Tm nhng im thuc trc tung qua c th k c 3 tip tuyn n th hs 1 22 4+ = x x ywww.VNMATH.com44 Cu 16) Tm nhng im thuc ng thng x=2 t k c 3 tip tuyn n th hs x x y 33 = Cu 17) Tm nhng im thuc trc Oy qua ch k c mt tip tuyn n th hs 11+=xxy Cu 18) Cho hm s 1 +=xm xya)Kho st v v th hm skhi m=1 b)Vi gi tr no ca m th hm s ct ng thng y=2x+1 ti 2 im phn bit sao cho cc tip tuyn vi th ti 2 im song song vi nhau. Phn ba: CC BI TON TNG GIAO 2 TH Cu 1) Cho hm s 2 2 2 34 ) 1 4 ( 2 m x m mx y + =a)Kho st v v th hm skhi m=1 b)Tm m th hs tip xc vi trc Ox Cu 2) Cho hm s 2 3 2 42 m m mx x y + =a)Kho st v v th hm skhi m=1 b)Tm m th hs tip xc vi trc Ox ti 2 im phn bit Cu 3) Cho hm s 253224+ = xxya)Kho st v v th hm s b)Tm phng trnhsau c 8 nghim phn bitm m x x 2 5 62 2 4 = + Cu 4) Cho hm smx mx x y 6 32 3 =a)Kho st v v th hm skhi m=1/4 b)Bin lun s nghim0 4 6 3 423= a x x xCu 5) Cho hm sx x y 3 43 =(C ) a)Kho st v v th hm s (C ) www.VNMATH.com45 b)Tm m phng trnhm m x x 4 4 3 43 3 = c 4 nghim phn bit Cu 6) Cho hm s) 1 ( ) 1 ( 3 32 2 2 3 + = m x m mx x ya)Kho st v v th hm s khi m= 1 b)Tm m hm sct Ox ti 3 im phn bit c honh dng Cu 7) Cho hm s) 5 ( 2 ) 7 5 ( ) 2 1 ( 22 3+ + + + = m x m x m x ya)Kho st v v th hm s khi m= 5/7 b)Tm m th hs ct Ox ti 3 im c honh nh hn 1. Cu 8) Tm m th hsm m x m m mx x y + + =2 2 39 ) 4 ( 2 3ct trc Ox ti 3 im to thnh 1 cp s cng Cu 9) Tm m hm s 8 ) 4 5 ( ) 1 3 (2 3 + + + = x m x m x yct Ox ti 3 im lp thnh cp snhn Cu 10) Tm m hm s1 2 ) 1 ( 22 4+ + + = m x m x yCt Ox ti 4 im to thnh cp s cng Cu 11) Chng minh rng th hs 11 2=xxyc 2 trc i xng Cu 12) Tm m hm s 8 18 ) 3 ( 3 22 3 + + = mx x m x yc th tip xc vi trc Ox Cu 13) Cho hm s2 32 4+ + = x x ya)Kho st v v th hs b)Bin lun s nghim phng trnhm x x = ) 1 ( 22 2 Cu 14) Cho hm s3 32 3+ + = x x x ya)Kho st v v th hm sb)Bin lun theo m s nghim phng trnh1 2 )33( 12+ =+ mxxPhn bn: CC BI TON LIN QUAN N KHONG CCH Cu 1) Tm M thuc (H) 25 3=xxy tng khong ccht M n 2 ng tim cn ca H l nh nht www.VNMATH.com46 Cu 2) Tm M thuc (H) :11+=xxy tng khong ccht M n 2 trc to l nh nht Cu 3) Tm trn mi nhnh ca th hm s (H): 39 4=xxycc im M1, M2 2 1M Mnh nhtCu 4) Tm trn mi nhnh ca th hm s 15 22 + =xx xycc im M, N di MN nh nhtCu 5) Tm trn th hm s 12 22 +=xx xyim M sao cho MI nh nht vi I l giao im 2 ng tim cnCu 6) Tm m hm sy=-x+m ct th hm s 21 2++=xxyti 2 im A,B m di AB nh nht MT S DNG BITP TNG HPKHC Cu 1) Cho hm s 4 22 1 y x mx m = + (1) , vim l tham s thc. 1)Kho st s bin thin v v th hm s (1) khi1 m = . 2)Xc nhm hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c din tch bng4 2 . Cu 2) Cho hm s 4 22 1 y x mx m = + (1) , vim l tham s thc. 1)Kho st s bin thin v v th hm s (1) khi1 m = . 2)Xc nhm hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c bn knh ng trn ngoi tip bng 1. Cu 3) Cho hm s 4 2 22 y x mx m m = + + +(1) , vim l tham s thc. 1)Kho st s bin thin v v th hm s (1) khi2 m = . 2) Xc nhm hm s (1) c ba im cc tr, ng thi cc im cc tr ca th to thnh mt tam gic c gc bng120. Cu 4) Cho hm s 4 22 y x mx = (1), vim l tham s thc. 1)Kho st s bin thin v v th ca hm s (1) khi1 m = . 2)Tmm th hm s (1) c hai im cc tiu v hnh phng gii hn bi th hm s v ng thng i qua hai im cc tiu y c din tch bng 1. Cu 5) Cho hm s( ) ( )4 2 22 2 5 5 y f x x m x m m = = + + +1/ Kho st s bin thin v v th(C ) hm s vim = 1 2/ Tm cc gi tr ca m thhm s c cc im cc i, cc tiu to thnh mt tam gic vung cn.Cu 6) Cho hm s 3 212 33y x x x = +(1) www.VNMATH.com47 1).Kho st s bin thin v v th ca hm s (1) . 2)Gi, AB ln lt l cc im cc i, cc tiu ca th hm s (1). Tm imMthuc trc honh sao cho tam gicMABc din tch bng 2.Cu 7) Cho hm s 3 26 9 4 y x x x = + (1) 1)Kho st s bin thin v v th ca hm s (1) 2)Xc nhksao cho tn ti hai tip tuyn ca th hm s (1) c cng h s gck . Gi hai tip im l 1 2, M M . Vit phng trnh ng thng qua 1Mv 2Mtheok . Cu 8) Cho hm s 3 23 4 y x x = + (1) 1.Kho st s bin thin v v th (C) ca hm s (1) 2. Gi s, , A B Cl ba im thng hng thuc th (C), tip tuyn vi (C) ti, , A B Ctng ng ct li (C) ti ' ' ', , ABC . Chng minh rng ba im ' ' ', , ABCthng hng. Cu 9) Cho hm s 33 1 y x x = +(1)1)Kho st s bin thin v v th (C) ca hm s (1). 2)ng thng ( A ):1 y mx = +ct (C) ti ba im. Gi A v B l hai im c honh khc 0 trong ba im ni trn; gi D l im cc tiu ca (C). Tmm gc ADB l gc vung. Cu 10) Cho hm s ( )3 2 2 23 3 1 3 1 y x x m x m = + + (1), vim l tham s thc. 1.Kho st s bin thin v v th ca hm s (1) khi1 m = . 2. Tmm hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cng vi gc to O to thnh mt tam gic vung tiO. Cu 11) Cho hm s( ) ( )22 2 1 y x x = (1) 1.Kho st s bin thin v v th (C) ca hm s (1). 2.Tmm th (C) c hai tip tuyn song song vi ng thngy mx = . Gi s, MNl cc tip im. Hy chng minh rng trung im ca on thngMNl mt im c nh (khim bin thin) Cu 12) Cho hm s 3 23 4 y x x = +(1)1)Kho st s bin thin v v th (C) ca hm s (1). 2)Gi kdl ng thng i qua im( ) 1; 0 Avi h s gck ( ) k R e . Tmk ng thngkdct th (C) ti ba im phn bit v hai giao im, B C ( BvCkhcA ) cng vi gc to O to thnh mt tam gic c din tch bng 1. Cu 13) Cho hm s 3 23 4 y x x = +(1) 1)Kho st s bin thin v v th (C) ca hm s (1). 2)Cho im( ) 1; 0 I . Xc nh gi tr ca tham s thcm ng thng: d y mx m = +ct th (C) ti ba im phn bit, , I A B sao cho2 2 AB < . Cu 14) Cho hm s:3 2 2 22( 1) ( 4 1) 2( 1) y x m x m m x m = + + + +1.Kho st v v th hm s khi m=0 2.Tm m hm s c cc tr , ng thi cc im cc tr 1 2; x x tho mn : 1 21 21 1 1( )2x xx x+ = +Cu 15) Cho hm sy = 2x3 + 9mx2 + 12m2x + 1, trong m l tham s. 1)Kho st s bin thin v v th ca hm s cho khi m = - 1. www.VNMATH.com48 2)Tm tt c cc gi tr ca m hm s c cc i ti xC, cc tiu ti xCT tha mn: x2C= xCT. Cu 16 Cho hm s3 2y (m 2)x 3x mx 5 = + + + , m l tham s1)Kho st s bin thin v v th (C ) ca hm s khi m = 0 2)Tm cc gi tr cam cc im cc i, cc tiu ca thhm s cho c honh l cc s dng. Cu 17) Cho hm s 2 12xyx+= (1) 1.Kho st s bin thin v v th( ) H ca hm s (1) . 2.Chng minh rng th( ) Hc v s cp tip tuyn song song, ng thi cc ng thng ni tip im ca cc cp tip tuyn ny lun i qua mt im c nh. Cu 18) Cho hm s( )xxx f+=11 2 ( H ) 1/ Kho st s bin thin v v th (H) ca hm s 2/ Gi () l tip tuyn ti im M( 0; 1 ) vi th (H). Hy tm trn (H) nhng im c honh x > 1 m khong cch t n () l ngn nht. Cu 19) Cho hm s 2m xyx=+ (Hm). Tm m ng thng d:2x+2y-1=0 ct (Hm) ti 2 im phn bit A, B sao cho tam gic OAB c din tch bng 38 Cu 20) Cho hm s 2 32xyx+=+. Tm nhng im M thuc th sao cho tip tuyn ti M ct haitim cn ti A, B sao cho vng trn ngoi tip tam gic IAB c bn knh nh nht. Vi I l giao im ca hai ng tim cn Cu 21) Tm m hm s 32 y x mx = +ct Ox ti mt im duy nht Cu 22) Cho hm s 2 12xyx+= (C). Tm hai im M, N thuc (C) sao cho tip tuyn ti M, N song song vi nhau v khong cch gia hai tip tuyn l ln nhtCu 23) Cho hm s 2 41xyx+= (H). Gi d l ng thng c h s gc k iqua M(1;1). Tm k d ct (H) ti A, B m3 10 AB =Cu 24) Tm m th hm s 3 22 y x mx m = +ct trc Ox ti mt im duy nht Cu 25) Cho hm s: 21xyx+=(C) 1) Kho st v v th (C) hm s2) Cho im A( 0; a) Tm a t A k c 2 tip tuynti th (C) sao cho 2 tip im tng ng nm v 2 pha ca trc honh Cu 26) Cho hm s 33 2 y x x = +(C)1) Kho st v v th hm s (C) 2) Tm im M thuc (C) sao cho tip tuyn ti M ct (C) N m2 6 MN =www.VNMATH.com49 Cu 27) Cho hm s 2( )m xy Hx m=+ v A(0;1) 1) Kho st v v th hm s khi m=1 2) Gi I l giao im ca 2 ng tim cn . Tm m trn th tn ti im B sao cho tam gic IAB vung cn ti A. Cu 28) Cho hm s 4 22 y x x = (C) 1) Kho st v v th hm s 2) Ly trn th hai im A, B c honh ln lt l a, b.Tm iu kin a v b tip tuyn ti A v B song song vi nhau. Cu 29) Cho hm s 22 2xyx+= (H) 1) Kho st v v th hm s (H). 2) Tm m ng thng (d): y=x+m ct th hm s (H) ti hai im phn bit A, B sao cho 2 2372OA OB + =Cu 30) Cho hm s y =3 22 (1 ) y x x mx m = + +(1), m l tham s thc. 1. Kho st s bin thin v v th ca hm s khi m = 1. 2. Tm m th ca hm s (1) ct trc honh ti 3 im phn bit c honh 1 2 3; ; x x x tho mn iu kin 2 2 21 2 34 x x x + +