Ôn luyện theo cấu trúc đề thi môn Toán Phần 1 Ôn thi tốt nghiệp THPT TS.Vũ Thế Hựu, Nguyễn Vĩnh Cận

510.76 6-454L d TS VO THE HlTU - NGUYEN VINH CAN TIICO C n U TRUC DC Till MON TS VU THE HlTU - NGUYEN VINH CAN ON LUVEN m THEO cAu TDUC DEI THI M6]\N ON THI TOTNGHIEP THPT QUOC GIA (2 1) lUHA XUAT BAN DAI HOC QUOC GIA HA NOI GIJII TICH Hoc va On luy§n theo CTDT mfln Toan THPT E I IK tX yYENflEl.fllllSOTOHIfPVflXIJCSU'' §1 HOAN VI, CHINH HdP, TO HOP KIEN THLTC Quy tac C Q n g va quy t ^ c n h a n a) Quy tdc cong : Neu tap hc(p A c6 m phan tuT, tap hop B c6 n phan tuf va giufa A va B khong CO phan tuf chung t h i c6 m + n each chon mot phan tuf thuoc A hoac thuoc B b) Quy tdc nhdn : De hoan mot cong viec A phai thuc hien hai cong doan Cong doan I CO m each thifc hien, cong doan I I c6 n each thiTc hien t h i c6 m n each de hoan cong viec A Tong quat, de hoan cong viec A phai qua k cong doan Cong doan thuf i (1 < i < k) c6 m i each t h i t h i c6 m i m m k each de hoan cong viec A Hoan vi Mot tap hop A hufu han c6 n phan tuf (n > 1) Moi each sap thuf tii cac phan tuf cua tap hop A dage goi la mot hoan vi cua n phan tuf cua A Dinh li : So hoan v i khac cua n phan tuf bang : P„ = n(n - l)(n - 2) 2.1 = n! C h i n h hofp Mot tap hop A hufu han gom n phan tijf (n > 1) va so nguyen k (0 < k < n) Moi tap hop cua A gom k phan tuf sap theo mot thuf t i i nhat dinh dLfOe goi la mot chinh hop chap k cua n phan tuf Dinh li : So chinh hop chap k cua n phan tuf bang : A;; = n ( n - l ) ( n - ) ( n - k + l ) = - ^, (n-k)! (Quy \x6c : 0! = 1) Tohtfp Cho tap hop A hufu han c6 n phan tuf (n > 1) va so' nguyen k (0 < k < n) Moi tap hop eon gom k phan tuf eiia A (khong tinh thuf tiJ cac phan tuf) goi la mot to hap chap k cua n phan tuf n! - A'' Binh li : So to hop chap k cua n phan tuf la : Cj^ =( n - k ) ! k ! k ! Hequd: Cl^Cl=l; ik-i C); = Cr" ; C);,, = C^ + C^ HQC fln luy?n theo CTDT mOn ToSn THPT BAI TAP Cho cac chff so 2, 3, 4, 5, 6, a) Co bao n h i e u so tir n h i e n c6 h a i chiJ so dirge tao n e n tCr t a p hgp eae chCr so da eho b) Co bao n h i e u so tir n h i e n c6 h a i ehuf so khac dugc tao n e n ttf tap hgj) cac chiJ so da cho CHI DAN a) b) De tao m o t so c6 h a i chuf so t a thiTe h i e n h a i eong doan : Chgn m o t chuf so l a m ehuf so h a n g ehuc : eo k e t qua c6 the Chgn m o t chuf so l a m ehuf so h a n g don v i : eo k e t qua c6 the Theo quy t&c n h a n so' k e t qua tao t h a n h cac so' c6 h a i ehuf so tCr tap hgp chuf so da eho la : n = x = 36 so L a p luan giong n h u eau a) n h i m g liTu y sir khac biet so vdi trirdng hgp t r e n d cho so dirge tao t h a n h eo h a i chCT so khac Do t a c6 k e t qua nhif sau : Chgn m o t chuf so l a m ehOf so h a n g ehuc : eo k e t qua eo t h e Chgn m o t ehuf so l a m chuf so hang don v i : eo k e t qua eo the (vi chQ so p h a i khac chij; so h a n g ehue da ehgn triTdrc do) Theo quy tie n h a n : so eae so eo h a i chuf so khac dirge tao t h a n h tCr t a p hgp eha so da cho la : n' = x = so Cdch khac : M i so eo h a i ehiJ so tao t h a n h tCr chuf so da eho la mot tap hgp sap thuf t i i gom h a i p h a n tuf tu" p h a n tuf da cho Do so cac so C O h a i chuf so khac tao t h a n h tCr ehuf so da eho la so e h i n h hgp chap cua tap hgp p h a n tuf n = = 6.5 = so' Cho t a p hgp cac chuf so 0, 1, 2, 3, 4, 5, a) Co bao nhieu so tir n h i e n c6 chuf so tir tap hgp cac ehuf so da eho b) Co bao n h i e u so tiT n h i e n eo ehuf so khac tiTng doi tiT tap hgp eae chuf so da eho C H I DAN a) Co each ehgn chuf so hang n g h i n (chuf so dau t i e n phai khac 0), each ehgn chiT so' h a n g t r a m , each chgn chuf so' h a n g ehuc va each ehgn ehuf so h a n g don v i Theo quy t^c n h a n : so each tao t h a n h so tiT n h i e n chuf so tCr t a p hgp chij" so da eho l a : N = x x x = 2058 so b) Co each chgn chuf so' h a n g n g h i n , k h i ehgn xong ehC? so' h a n g n g h i n l a i chOr so khae v d i ehuf so h a n g n g h i n da chgn Vay c6 each chgn ehuf so h a n g t r a m K h i da chgn chiT so h a n g n g h i n va hang t r a m , eon l a i ehu: so khac v d i cac chuT so da chgn Do eo each a TS Vu Thg' Huu - Nguyen Wnh C?n chon chOf so h a n g chuc Ttfang t i i , c6 each chon chOf so h a n g dcfn v i Theo quy tac n h a n So cac so t u n h i e n c6 chijr so khac tCrng doi ducfc tao t h a n h til t a p hop chOf so da cho l a : N ' = X X X = so Cdch lap luan khdc : M o i so t i i n h i e n c6 chOr so khac tao t h a n h tCr t a p hop chCf so da cho l a m o t c h i n h h o p chap tCr t a p h o p chuf so ma cac c h i n h hop n a y k h o n g c6 chOf so' or dau Do so' cac so C O chOr so khac til chuf so l a : N' = = X X X - X X = so Mot to hoc sinh c6 10 ngufofi xep thCf td hang de vao Icfp H o i a) Co bao n h i e u each de to xep h a n g vao Idfp b) Co bao nhieu each de to xep h a n g vao Idp eho h a i b a n A va B cua to luon d i canh va A dufng triTdfc B CHI DAN a) So' each xep h a n g bang so' hoan v i cua 10 p h a n tuf N i = 10! = 3628800 each b) Coi h a i b a n A va B n h i i m o t ngifdi Do so each xep h a n g cua t o de vao Idp t r o n g h a i b a n A va B d i l i e n n h a u bang so h o a n v i cua p h a n tuf N2 = 9! = 362880 each Co bao nhieu each xep ngUc/i n g o i vao m o t b a n a n cho t r o n g cac trirdng hop sau : a) Sip ngiidi theo h a n g ngang cua m o t b a n a n d a i b) Sip ngLfcfi ngoi vong quanh m o t b a n a n t r o n CHI DAN a) M o i each ngoi theo h a n g ngang l a m o t hoan v i cua p h a n tuf So each sap xep la : 6! = 720 each b) Gia sijf ngiicfi a n dirge d a n h so thuf t u l a : 1, 2, 3, 4, 5, v a m o t each sap xep theo b a n t r o n nhiT h i n h /^^^T^K^ ^ \ ^ _ _ ^ j (1) (2) ^\ J Neu thi dai ban (3) (4) (5) (6) t a cat b a n t r o n d v i t r i giufa v a r o i t r a i d a i theo b a n ngang t a C O hoan v i (1) tiTcfng lirng m o t each xep ngudi n g o i theo b a n a n Tirong t i i cat d v i t r i giufa va N h i i vay m o t each sSp xep theo t r o n tifOng ufng v d i each sap xep theo b a n d a i Do so each Hpc v4 On luy$n theo CTDT mfln Toan THPT £3 xep ngiicri ngoi quanh b a n a n t r o n l a : N = — = 120 each M o t to CO 15 ngiJcfi gom n a m va niJ C a n lap n h o m cong tac c6 ngiTcfi H o i c6 bao n h i e u each t h a n h l a p n h o m t r o n g m o i triTofng hop sau day : a) N h o m c6 n a m va nvC b) So n a m va nOf t r o n g n h o m bSng c) P h a i CO i t n h a t m o t n a m CHI D A N 87 a) So each chon n a m t r o n g so n a m l a : Cg = " = 84 1.2.3 So each chon nuf t r o n g so nuf la : Cg = So each lap nhom gom nam va nuf (theo quy tac nhan) la : Ni = C^C^ = 504 each b) So each l a p n h o m gom n a m va nuf la : N = C ^ C ^ = — — = 540 each ' ' 1.2 1.2 c) So each t h a n h l a p n h o m ngudi t r o n g c6 i t n h a t n a m la : n a m , nuf hoSe n a m , nOr hoSc n a m , nuf hoSc n a m N = C^.C^+C^C^+C^C^+C^ ^ 6.5.4 9.8 6.5 9.8.7 ^ 9.8.7.6 = + + + = 1350 each 1.2.3 1.2 1.2 1.2.3 1.2.3.4 Ghi chu : Cung c6 the l a p l u a n nhiJ sau : Ca to CO 15 ngUdi So each l a p n h o m ngUcJi t u y y l a : ^4 15.14.13.12 , , Ci5 = — — = 1365 each 1.2.3.4 So each l a p n h o m ngudi t o a n nuf la : Cg = Cg = — ^ = 15 each So each l a p n h o m ngiTdi c6 i t n h a t n a m la : N = C^5 - C^ = 1365 - 15 = 1350 each T r o n g mSt p h a n g c6 n d i e m p h a n b i e t ( n > 3) t r o n g c6 dung k d i e m n k m t r e n m o t d i T d n g t h a n g (3 < k < n ) H o i c6 bao n h i e u t a m giae n h a n eac d i e m da cho la d i n h CHI D A N C i i d i e m k h o n g t h a n g h a n g tao t h a n h m o t t a m giae So eac t a p hdp d i e m t r o n g n d i e m la : C^ So eac t a p d i e m t r o n g k diem t r e n dirdng t h a n g la : C^ So t a m giae c6 d i n h la eac d i e m da cho l a : N = C ^ - C ^ t a m giae £ TS Vu ThS' Hi/u - Nguyen V i n h C?n a) Co bao n h i e u so tu n h i e n l a so chSn c6 chOr so doi m o t khac v a chiJ so dau t i e n l a chui so le b) Co bao nhieu so tuT n h i e n c6 chuf so doi m o t khac nhau, c6 dung chCif so le, chuf so chSn (chuf so dau t i e n phai khac 0) CHI D A N a) So can t i m c6 dang : so 0, 1, 2, 8, vdi x = ajagaga^Hgag aj ;t 0, aj vdi t r o n g 1< i ai, ae l a y cac chijr ;t j < - V i X la so chSn nen ag c6 each chon tii cac chuf so 0, 2, 4, 6, - V i a i la chuf so le nen c6 each chon tif cac chuf so 1, 3, 5, 7, Con l a i a a a a l a m o t c h i n h hop chap cua chuf so l a i sau k h i da chon ae va a i Theo quy tSc n h a n , so cac so can xac d i n h l a : N i = 5.5.Ag = 5.5.8.7.6.5 = 42000 so b) M o t so theo yeu cau de b a i gom chuf so tCr t a p X i = {0; 2; 4; 6; 81 va chuf so tii t a p hop X2 = { ; 3; 5; 7; 9) ghep l a i va l o a i d i cac day chuf so CO chuf so dufng dau So each l a y chuf so thuoc t a p X i l a : C = 10 each So each l a y p h a n tijf thuoc X2 l a : C = 10 each So each ghep p h a n tuf l a y tii X i v d i p h a n tuf l a y tii X2 l a : C^.C^ = 10.10 = 100 each So day so eo thuf tir cua p h a n tuf diicfc ghep l a i l a : 100.6! = 72000 day Cac day so c6 chuf so of dau gom chuf so khac cua X i v a chijf so cua X2 : So cac day so n h u t r e n l a : C ' C g S ! = 7200 day So cac so theo yeu cau de b a i l a : N2 = C ^ C ^ ! - C ^ C ^ ! = 72000 - 7200 = 64800 so M o t hop diing v i e n b i do, v i e n b i t r S n g va v i e n b i v a n g NgUcfi ta chon r a v i e n b i tii hop H o i c6 bao n h i e u each l a y de t r o n g so b i l a y r a k h o n g du ca m a u CHI D A N Cdch : So each chon v i e n b i k h o n g du m a u b a n g so each chon v i e n b a t k i trCr d i so each chon v i e n eo ca m a u N = C\, -{ClC\.C\+Cl.C\.C\+Cl.C\.C\) = 645 each Cdch : So each chon v i e n b i k h o n g du m a u b a n g so each chon v i e n m o t m a u (4 do, t r a n g va vang) eong v d i so each chon v i e n h a i m a u (1 do, t r a n g hoac do, t r a n g hoac do, t r S n g hoac do, v a n g hoSe do, v a n g hoac do, v a n g hoSc t r a n g , v a n g hoac t r a n g , v a n g hoac t r a n g , vang) N=C^ + + + C^C^ + ClCl + C^C^ + + C^C^ = 645 each Hoc va On luyen theo CTBT mfln ToSn THPT S DAN CHI Co 15 n a m va 15 nuf khach du l i c h duTng t h a n h vong t r o n quanh ngon lufa t r a i H o i c6 bao nhieu each xep de k h o n g c6 triTdng hop h a i ngiTdi cung g i d i canh ThiTc h i e n sAp xep b k n g each danh so" 30 cho t r e n diTdng t r o n tii den 30 va cho n a m duTng so le nuf dufng cho so chkn hoSc ngiioc l a i (2 each) Co 15! each sSp n a m dufng t r o n g eae ch6 so le (hoac ehSn) va 15! each sap nuf dufng t r o n g cac eho so ehSn (hoftc le) V i dudng t r o n 30 cho n e n m o i each sap xep nao xoay tua 30 eho theo diing t r a t tiT t a cung chi c6 m o t each sap t r e n ducfng t r o n (xem b a i so 4) Do so each sSp xep theo diiofng t r o n 30 k h a c h du l i c h theo yeu cau „ 2.(15!)(15!) ^ , , de la : N = = 14!.15! each 30 10 Chufng m i n h cac dSng thuTc : 1 = ——- (1) t r o n g A^ la c h i n h hop chap cua n a) -ir + —7r + - + A„ n b) c;; = c;;:'i + c;;:'^ + + c;::; (2) c;; la to hop chap r cua n CHI DAN a) V d i k e N , k > t a eo : A^ = k ( k - 1) J_ k(k -1) k-1 - ~ (*) T h a y k = 2, 3, n vao (*) ta c6 ve t r a i cua (1) l a : (1 1^ (1 1^ _^ _ n-1 ^ + — + + ( — n n 2j ^n-l nj U 3j b) Theo t i n h chat cua to hop ta c6 : u lr-l Cn-2 = Cn.3 + C;;_3 Cong ve v d i ve eae dang thufc t r e n t a dirgfc : c : ; = c - + C - U C - + + C - + C : Do C[ = C[:\ n e n thay C; d dSng thufc euoi boi C^:} t a dUdc dang thiJc (2) can chufng m i n h 11 Chufng m i n h b a t dSng thufc : C^QOI + C'OM ^ C^^i + t r o n g k e N , k < 2000, ^Zl la to hop chap k cua n phan tuf £ TS Vu ThS' Huu - Nguyen Vinh C$n 10 CHI DAN V d i < k < 1000 t h i CLi C ^ 2001! k!(2001-k)! pk ^ pk+1 ^ pk+2 '-^2001 - '-^2001 - ^ 0 (k + l ) ! ( 0 - k ) ! 2001! ^ ^ plOOO _ p l O O l - ••• - '-'2001 " '-^2001 k+ 2001-k ^ p k *-^2001 hay ax + b > 0; a, b e R, a ?i • N e u a > 0, ax + b > CO t a p n g h i e m x>- — a • N e u a < 0, ax + b > CO t a p n g h i e m x< — a D a u c u a n h i thufc b a c n h a t ax + b (a ;t 0) Nhi thufc bac n h a t : f = ax + b (a ^ 0) Gia t r i x = - — t a i f c6 gia t r i hkng goi l a n g h i e m cua n h i thufc a Ta CO d i n h 11 ve dau cua n h i thufc n h u sau : Dinh U + V6i cac gia tri x < - — t h i f = a x + b c dau t r a i v6i dau cua he so a a + V d i cac gia t r i , a) G i a i cac phuong t r i n h : 10-x 20-x 30-X = + 120 100 110 x > - — t h i f = ax + b c6 dau cung dau vcfi he so a a (1) Hoc va Sn luy?n theo CTDT mOn ToSn THPT S b) X - X - X + = c) x^ + x ' + 5x^2 + x - 12 = CHI DAN a) (1) o 10-X 100 -1 -90-X 100 X + 20-X 110 -90-X + 100 = « (2) (3) 30-X -1 120 -90-X + 110 (90 + x ) b ) x^ - x^ - o + 110 120 ^ 120 -1 -0 ^ = = Oci> X = -90 x^Cx - 1) - ( x - 1) = (x - l ) ( x ^ - 1) = x ' - l = o x - l = o X = ±1 c) x^ + 2x^ + 5x2 + x - 12 = o ( x ^ - ) + 2(x^ - ) + SCx^ - 1) + ( x - 1) = o ( x - l ) [ x ^ + Sx^ + x + 12] = 0 Vx) 1) G i a i , b i e n l u a n t h e o t h a m s o ' m phufong t r i n h : CHI DAN m \ m = 9x + m^ (1) ( I ) c : > ( m - ) x = m2 + m + N e u m ^ - ;t 0, n g h i a l a v i m X — + + m^ + m ±3 t h i phuTcfng t r i n h (1) c6 n g h i e m : m — m^ - m - N e u m = 3, p h u o n g t r i n h (1) c6 d a n g : x - = x + Phiicfng t r i n h n a y v6 n g h i e m N e u m = - t h i p h t f o n g t r i n h (1) c6 d a n g : x + = x + Phucfng t r i n h v so n g h i e m ( m o i so t h u c deu l a n g h i e m ciia p h i f o n g t r i n h ) C h o p h u o n g t r i n h : a^x + b = a x + ab (1) t r o n g a, b l a cac t h a m so t h i f c G i a i , b i e n l u a n t h e o a, b p h i / o n g trinh tren CHI DAN + ( l ) c : > a ( a - l ) x = b ( a - 1) N e u a(a - 1) 0, tufc l a n e u a ?t o v a a ;^ t h i (1) c6 n g h i e m d u y n h a t (a-l)b b X = + = a(a-l) — a N e u a = 0, p h u o n g t r i n h (1) c6 d a n g : Ox = - b T r o n g t r i r d n g h o p n a y c6 h a i k h a n a n g : 24 a N e u b = (tufc l a a = 0, b = 0) m o i so t h t f c d e u l a n g h i e m - N e u h ^ ( t i i c l a a = 0, b ?i 0) p h i i d n g t r i n h v n g h i e m - TS, Vu The Hi/u - Nguyen Vinh CSn + Neu a = 1, phtrcfng t r i n h (1) c6 dang : Ox = 0, phiicfng t r i n h c6 t a p n g h i e m la m o i so thuc 4, G i a i cac phifong t r i n h : a) (x - if + (x + 2f = (2x + 1)^ (1) b) 9ax^ - 18x^ - 4ax + = (2), a la t h a m so CHi DAN a) ( ) « (2x + l ) [ ( x - (x - I X x + 2) + (x + 2f] = (2x + 1)^ o (2x + l ) [ ( x - if - (x - l ) ( x + 2) + (x + 2f - (2x + 1)^] = (2x + l ) ( x + 2)(x - 1) = o b) (2) + + a) CHI a) b) x = X = - , x = ax(9x^ - 4) - 2(9x^ - 4) = (ax - 2)(3x + 2)(3x - 2) = 2 Neu a ?t 0, (2) c6 cac n g h i e m : x = —, x = ±— a Neu a = 0, (2) c6 cac n g h i e m : x = ± — G i a i cac bat phUcfng t r i n h dufdi day va bieu d i e n t a p n g h i e m cua no t r e n true so 3(x + 2) - > 2(x - 3) + b) (x - 2)^ + + x^ > 2^ - 3x - DAN -7 X c^3x + > x - 3x - 2x > - - X > - ^ Tap n g h i e m diTOc bieu d i e n h i n h ben inx a o x^ - 4x + + + x^ > 2x^ - 3x - o - x + 3x > - - o - x > - « x < ^ = Q -1 Bieu dien t a p n g h i e m b h i n h ben Hinh G i a i va b i e n l u a n theo t h a m so m b a t phirong t r i n h : x + _ CHI DAN — - > '^^"*"-(m + l ) x m m g ^ b (1) Dieu k i e n xac d i n h m ; t Chuyen ve, r u t gon t a diTOc : ( m + 2)x > — m + Neu m 5t va m > - ( t h i m + > 0) b a t phtfong t r i n h c6 n g h i e m X > m ( m + 2) + Neu m < - ( t a t n h i e n m ;^ 0) t h i (1) c6 t a p n g h i e m x < + Neu m = - bat phiJOng t r i n h t r d t h a n h Ox > - M o i so thuc deu la n g h i e m cua (1) m ( m9 + 2) G i a i , b i e n l u a n b a t phiTcfng t r i n h v d i t h a m so a va b: x + > — + — (1) b a HQC fln luy§n theo CTOT m5n Toan THPT S 25 CHI DAN (a^ - ab)x + - ab a-b, ,^ ^ , (1) ; — < — — ( a x - b) < (a 0, b 0) ab ab + N e u a = b ? i t h i (1) c6 dang Ox - < v6 n g h i e m + N e u a < b < hoSc < b < a t h i (1) c6 n g h i e m x < — a + N e u a < < b hoSc b < < a hoac b < a < hoSc < a < b t h i (1) c6 tap n g h i e m l a : x > — a G i a i cac b a t philcfng t r i n h : a) (2x - 3)(5 - x) > X + CHI DAN a) D u n g d i n h l i ve dau cua n h i thufc : ax + b (a 0) 2x - CO n g h i e m x = — Neu X < — t h i 2x - c6 dau a m ( t r a i dau v — t h i 2x - CO dau dtrong (cung dau vdi 2) Tuong tir - X > v d i X < va - X < v d i X > L a p bang xX e t dau cua t i c h (2x - 3)(5 - x) nhir sau : -co — 2x- - X + + + - +00 + - (2x - 3X5 - x) + Can cur bang x e t dau t a t h a y t a p n g h i e m cua b a t phildng t r i n h l a — < X < b) Dieu k i e n xac d i n h : x ^ - L a p bang x e t dau nhir cau a) : X 5 -5 -00 — — 2x- X + 2x-l X + 26 S TS Vu Thg' Huu - +CC - + + + - + 0 + Nguyin VTnh C§n a) CHI a) b) 10 Tap n g h i e m : - < x < — Giai cac b a t phiTcfng t r i n h : 3x-4 >2 b) x-2 x +2 DAN 3x-4 - > < o ^ ^ > o x+2 x+2 S S : X < - hoSc x > 11-x > o >0o x-2 x+1 (x - 2)(x + 1) DS :x 41 Tap n g h i e m cua he ( I ) l a : 19 X< — 11 • n •X 20 > 41 41x-20 ^ ^ 30 30 A 20 19 41 u 20 ^ 19 hay — < X < 41 G i a i b a t phuong t r i n h kep : x - < 2x + < x + (*) CHI D A N Bat phtfcrng t r i n h kep (*) la each v i e t khac cua he b a t phifong t r i n h sau JX- < 2x + (1) " x + < x + l (2) Bat phifong t r i n h (1) c6 t a p n g h i e m l a : x > - Bat phiiong t r i n h (2) c6 t a p n g h i e m l a : x < - Vay n g h i e m cua b a t phiiong t r i n h kep (*) l a : - < x < - HQC 6n luyen theo CTBT m6n Toan THPT £3 27 §2 PHUOfNG TRINH, BAT PHlICfNG TRINH BAC HAI MOT AN PhuToTng t r i n h b a c h a i m p t a n ax^ + bx + c = (a ?t 0) a) Cong thiic nghiem : Tong qudt : Biet thufc A = b^ - 4ac + Neu A < 0, (1) v6 nghiem + + + + + b) (1) Neu A = 0, (1) C O nghiem kep x = - — 2a Neu A > 0, (1) C O hai nghiem phan biet : _ -b-VA _ -b + V A ~ 2a ' " 2a ' Thu ggn : trudng hdp b = 2b', biet thufc thu gon A ' = b'^ - ac Ne'u A ' < 0, (1) v6 nghiem b' Neu A ' = 0, (1) C O nghiem kep x = a Neu A ' > 0, (1) C O hai nghiem phan biet : _ -b' - VA^ _ -b' + VA^ x, — , x, — a a Dinh li Vi-et : Neu Xi, X2 la hai nghiem cua phtfcrng trinh ax^ + bx + c = (a 0) t h i b = Xi + X2 = a P = X1X2 = - a Ung dung : Neu hai so c6 tong b&ng S va c6 tich bkng P t h i cac so la nghiem cua phuong t r i n h : x^ - Sx + P = fBAITAPJ Giai, bien luan theo tham so m phuong trinh : (m - D x ' - ( m + l)x + m = (1) 11 CHIDAN + Neu m - l = m = l , phuong trinh (1) la bac nhat, c6 nghiem nhat X = i + Neu m t l = > m - l t , ap dung cong thufc nghiem thu gon ta c6 : A' = (m + 1)^ - (m - Dm = 3m + + Neu m = — t h i A ' = 0, c6 nghiem kep x, = x„ = ^ = — 28 S] TS Vu The Hi^u - Nguyin VTnh Can + Neu m < — t h i A' < 0, v6 n g h i e m m ^1 m + + V3m + 1 t h i A' > 0, (1) CO h a i n g h i e m Xj = m >— • ' m-1 Vdti gia t r i nao cua t h a m so m cac phiTOng t r i n h cho durdi day cd n g h i e m chung 2x^ + m x - = (1) mx^ - x + = (2) + Neu 12 CHI D A N V d i m = 0, h a i phUdng t r i n h k h o n g cd n g h i e m chung Gia suf m 9t 0, b i e t thufc cua (1) l a : A i = m^ + > 0, V m , b i e t thuTc A2 cua (2) la : A2 = - m > v d i m < - Neu a l a n g h i e m chung t h i tCr (2) t a cd : a-2 ma^ - a + = 0=:>a = (3) m Thay vao (1) t h i dtroc : 2a^ + m a - = a-2 l-2m a' = m'+2' " m'+2 Thay vao (3) t a dUgfc he thijfc : a = m +4 l-2m ^m+4 m^ + 13 m'+2 ^ m'' + m + = m + m a - =0 => m = - 1, (Gia t r i m = - thda m a n dieu k i e n m < —) D a p so : m = - Cho phtfcfng t r i n h v d i t h a m s o ' m sau : x^ + ( m - l ) x + m - = (1) V d i gia t r i nao cija m phifcfng t r i n h cd cac n g h i e m x i , X2 t h o a m a n he thufc : x i + 3x2 = (*) CHI D A N B i e t thufc A = ( m - i f - ( m - 6) = m^ - 2 m + 25 T a can chpn gia t r i cua m cho A > de (1) cd n g h i e m X i , X2 Theo d i n h l i V i - e t : x i + X2 = - m TCr he thiic (*) t a cd : = x i + 3x2 X, ^ = xi + ( x i + X2) = 3(1 - m ) + x j = 3m - [(3m - 2f + ( m - l ) ( m - 2) + m - = => 12m(m - l ) = 0m = hoSc m = Cac gia t r i m = va m = t h d a m a n dieu k i e n A > V a y m = hoSc m = l a cac gia t r i can t i m Hue va Sn luy§n theo CTDT m6n Toan THPT S 14 Cho phtfcfng t r i n h wdi t h a m so m - 2(m + l)x + 2m + = (1) Xac d i n h cac gia t r i cua m de phifong t r i n h (1) c6 h a i n g h i e m X i , X2 cho bieu thiJc A = X j + X + l O X j X g d a t gia t r i n h o n h a t C H I D A N Gia suf (1) C O h a i n g h i e m x i , X2 t h i : A = xf + X2 + X X - (Xi + X ) ^ + 8X1X2 = ( m + 1? + 8(2m + 5) = ( m + 3)^ + A d a t m i n k h i m = - Gia t r i m = - thoa m a n dieu k i e n : A' = ( m + If - ( m + 5) > V a y m = - l a gia t r i can t i m 15 Cho phiiong t r i n h : x^ - 8x + 12 = (1) K h o n g can g i a i (1), h a y l a p m o t phuong t r i n h c6 h a i nghiem la n g h i c h dao cua cac n g h i e m cua (1) C H I D A N Goi cac n g h i e m cua (1) l a x i , X2 t h i cac n g h i e m cua phiTOng t r i n h can lap C O cac n g h i e m l a : — , — PhiTOng t r i n h can l a p la : x^- 1 +• X 4- = x^ i ^ X + = T h a y x i + X2 = 8, X1X2 = 12 t a difoc phifong t r i n h can t i m : X + — = X 16 12 T i m cac gia t r i cua t h a m so m t r o n g cac t r i f d n g h o p sau : a) x^ - 6x^ + m = (1) c6 n g h i e m p h a n b i e t b) x^ - ( m - l)x^ + m - = (2) c6 h a i n g h i e m p h a n b i e t C H I D A N a) Phuong t r i n h (1) c6 n g h i e m neu phifong t r i n h : - 6X + m = C O h a i n g h i e m dtfong Dieu n a y x a y r a neu 'A' = - m > •S = > o < m < P =m >0 b) Phuong t r i n h (2) c6 h a i n g h i e m p h a n b i e t k h i phifong t r i n h : X ' - 2(m - 1)X + m - = CO dung m o t n g h i e m diiOng DS : m < 3 m TS Vu ThS Hyu - Nguygn Vinh CSn §3 BAT PHlJCfNG TRINH BAG HAIMQT AN KIEN THCfC D a u c u a tarn thu!c h^c h a i f(x) = ax^ + bx + c (a 0) (1) a) Dinh li ve dau cua tarn thiic bac hai : Goi A = b^ - 4ac l a b i e t thufc cua tarn thurc (1) • N e u A < 0, Vx G R, f(x) c6 cung dau v d i a, n g h i a l a : aflx) > 0, V x G R • N e u A = t h i f(x) = v d i x = - — 2a va f i x ) cung dau v d i a, v d i m o i x ^ - — , nghia l a : aflx) > Vx e R, f i x ) = c:> x = - — 2a 2a • Neu A > t h i f(x) c6 hai nghiem phan biet x i , X2 (xi < X2), f(x) t r a i dau v i a v6i X e (xi; X2) va cung dau v i he so a vcri x G ( - C O ; xi) hoac x e (X2; + 0, Vx G ( - 0 ; xi) hoSc Vx G (X2; +00) b) Dinh li ddo ve dau cua tarn thiic bac hai Cho tarn thurc f(x) = ax^ + bx + c (a ^ 0) v a so a Neu afla) < t h i f(x) c6 h a i n g h i e m x i < X2 va x i < a < X2 He qua Dieu k i e n can v a du de phiJOng t r i n h : fix) = ax^ + bx + c = (a ^ 0) CO h a i n g h i e m p h a n b i e t la CO so a G R I a f I a ) < He qua Dieu k i e n can va du de phiTOng t r i n h : f(x) = ax^ + bx + c = CO h a i n g h i e m t r o n g c6 m o t n g h i e m t r o n g k h o a n g (a; P), a < p l a cac so cho trirdc, l a : f(a)f(p) < c) So sdnh mot so vai cdc nghiem cua tarn thiic f i x ) = ax^ + bx + c (a 5^ 0) • Ne'u afia) < t h i x i < a < X2 • N e u afIa) = t h i a l a n g h i e m • Neu afIa) > t a x e t t h e m A = b^ - 4ac i) N e u A < t h i f i x ) v6 n g h i e m , k h o n g so sanh ii) Neu A > va — ^ - a > t h i a < xi < 2a iii) Neu A > va - — - a < t h i 2a X i < X2 X2 < a BAI TAP 17 X e t dau cua f i x ) = — theo su b i e n t h i e n cua x t r e n R x ' - x + 12 Hoc ya an luy$n theo CTDT mfln Toin THPT S 31 ... -1 -90-X 10 0 X + 20-X 11 0 -90-X + 10 0 = « (2) (3) 30-X -1 120 -90-X + 11 0 (90 + x ) b ) x^ - x^ - o + 11 0 12 0 ^ 12 0 -1 -0 ^ = = Oci> X = -90 x^Cx - 1) - ( x - 1) = (x - l ) ( x ^ - 1) = x '' -... < 10 00 t h i CLi C ^ 20 01! k!(20 01- k)! pk ^ pk +1 ^ pk+2 ''-^20 01 - ''-^20 01 - ^ 0 (k + l ) ! ( 0 - k ) ! 20 01! ^ ^ plOOO _ p l O O l - ••• - ''-''20 01 " ''-^20 01 k+ 20 01- k ^ p k *-^20 01