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PGS TS LeVan Hi~n (Chu bien) Phan Van Toan Tong Van Phuc cao Van Phi Nguyen Quoc san Ninh Thi Hong Nguyen Do Chien 51 qu ~et chinh phL J c ~ NHA XUAT BAN Dl 1 HQC QUOC GIA HA NQI PGS TS LeVan Hi~n (C.
PGS.TS LeVan Hi~n (Chu bien) Phan Van Toan -Tong Van Phuc - cao Van Phi - Nguyen Quoc san - Ninh Thi Hong - Nguyen Do Chien 51 qu ~et chinh phL.J c ~· NHA XUAT BAN Dl:\1 HQC QUOC GIA HA NQI T939 I I •···························0···························• Taan hQC Ia mot trang nhll'ng man khaa hQC tl)' nhien nam trang chU'O'ng trlnh giaa dl)C thong (y ViE;)t Nam tCr b~c Tieu hQc d~n THPT Ve m~t ki~n thCrc, Taan hQc cung cap cha cac em cong Cl) de tinh taan va ti~n hanh nghien eCru cac man hQC khac nhU' V~t li, H6a hQC, Ve m~t kT nang, hQC taan giup cac em rem luyE;)n kha nang tU' duy, suy lu~n 16-gic, suy nghT m9ch l9c dong thai gop ph§n hlnh va phat trien kT nang giai quy~t van de trang CUQC song Vai d~c thu man hQC doi hoi c6 nen tang ki~n th(rc va kT nang th~t vCI'ng chac, chung toi luon tran tr& tlm nhll'ng phU'O'ng phap toi U'U nhat giup cac em ti~p c~n va hoc t~p vai taan hoc 8U'Q'C due rut tCr kinh nghiE;)m giang d9Y va nghien CLPU cac an pham ca trang va ngaai nU'aC, cuon sach Bi quyet chinh ph~;~c di~m cao Toan 9- T~p dU'Q'c chung toi thi~t k~ nham ho trQ' toi da cac thay co va cac em hoc sinh trang qua trlnh hoc t~p, on luyE;)n, kiem tra va danh gia ki~n thCrc HE;) thong ki~n th(rc bam sat chU'O'ng trlnh hoc tren lap •.••••• ·~ Ml)C tieu r6 rang thea tCrng bai -~ Thea sat ho trQ' cac em hoc t~p tren lap, on t~p, kiem tra danh gia kip thai ····~ Giup cac em c6 cai nhln tong quat, d~t mt:~c tieu de d9t dU'Q'c trU'ac hoc Trinh bay ki~n thCrc chi tiet, li giai day du giup cac em hieu sau sau d6 hE;) thong h6a l9i de cac em thay ro dU'Q'c sl)' logic cua kien thCPc, nhll'ng kien thCrc tam can nha Lf thuyet trlnh bay chi ti~t, k~t hQ'p SO' hE;) thong h6a l9i ki~n thCPc Bai t~p dU'Q'c phan d9ng chi ti~t cac bai taan, kieu cau hoi thU'ang g~p, c6 phU'O'ng phap giai chi tiet thea tCrng bU'ac va cac vi dl) minh hQa HE;) thong cac mE?a giai, kinh nghiE;)m lam bai, loi sai can tranh, kien th(rc bo sung va cac chu y di kern vai cac vi dl) HE;) thong de kiem tra 15 phut, 45 phut, hQC kl bam sat thea qua trlnh hQC Giup cac em nh~n biet dU'Q'c tCrng d9ng bai, kieu hoi va each tU' de giai d9ng bai, kieu hoi ·········~ ·········~ ··~ Truyen thl) cha cac em nhll'ng kinh nghiE;)m cua cac tac gia nhieu nam d9y hQC, giup cac em nam vCrng kien th(rc mai, on t~p l9i kien thCPc cO can ···~ Ho trQ' cac em on t~p dung thai diem de tl)' tin lam cac bai kiem tra tren lap Cau true Cl) the cua cuon sach nhU' sau: Ph&n 1: N{)i dung bai hQc Trang d6 phan dU'Q'C chia cac n()i dung l&n sau £)~i ChU'ang H~ phU'ang trlnh b~c nhat hai an so ChU'ang Ham so y = ax (a -:f ) PhU'ang trlnh b~c hai m()t an Hinh hQC ChU'ang G6c v&i dU'ong tron ChU'ang Hlnh trl) - hlnh n6n - hlnh cau Phan 2: a~ ki~m tra Phan 3: aap an V&i nhCPng U'U diem cua cuon sach, chung toi mang rang Bi quyet chinh phi:JC di~m cao Toan T~p se dem den cha cac em hQC sinh ngufln cam hling hQC t~p, dflng hanh cung cac em trang hanh trlnh chinh phl)C tri thlic va Ia ngu6n tai li~u tham khaa hCPu fch cha cac b9n d6ng nghi~p trang qua trlnh giang d9y Trang qua trlnh bien sa9n, chung toi khong the tranh kh6i nhCPng thieu s6t khong mang muon, rat mang se nh~n dU'Q'C nhCPng d6ng g6p quy bau tCr b9n dQC gan xa, cac thay co cQng nhU' cac em hQC sinh than men de chung toi haan thi~n cuon sach han trang nhCPng lan tai ban tiep thea Xin chan cam O'n! ? N M o Sau thai gian dai no ll)'C trien khai, trU'&C Sl)' ch1EM) I 1-A I 2-D I 3-C 4-C 5-C PHAN Tlf LU~N (7,0 £>1EM) cAu NQI DUNG BIEU £>1EM a) Ta c6 bang gia tri X y=-X2 -2 -1 2 2 o,5 diem E>6 !hi ham s& y = a(-1;i} ix di~m 18 parabol (P) di qua cac 0(0;0); A(1; ~} C(2;2) va 0(-2;2) y_., (P) D -~ C o,5 diem - B - - I A I I -2 Cau I I -1 b) PhU'crng trlnh hoanh de) giao diem cua !2 x x (P) va (d) Ia =- 2x- m x + 4x +2m= Ta c6 11' = 22 -2m = -2m 0,5 diem De (d) cat (P) t~i hai diem phan bi~t th1 11' > 4- 2m> m < V~y v&i m < th1 dU'ong thang d cat (P) t~i ~ai diem phan bi~t (P) va d Ia m=~, tac6 (d):y=-2x-~ Khi 2 PhU'crng trlnh hoanh de) giao diem cua -X o,5 diem =-2X X +4X+3=0 Ta c6 a - b + c = - + = PhU'ang tri n h c6 hai nghi$m phan bi$1 x, = -1 => Y1 = V~y lQa dQ hai giao di~m cUa (d) va (P) 18 M( i; -1;iJ X2 = -3 => Y2 = ~ va N( o,5 diem -3;~} a) Khi m = 5, ta dU'Q'C phU'crng trlnh 2x2 -4x+2 = x -2x+1 = (x-1t =0 x-1 = x =1 V~y v&i m = phU'crng trlnh da cho c6 nghi~m x = o,5 diem t b) Ta c6 11' = (-2 -2 (m - 3) = -2m + 10 De phU'crng trlnh c6 hai nghi~m phan bi~t th1 Cau 11' > -2m+ 10 > m < V~y v&i m < th1 phU'crng trlnh da cho c6 hai nghi~m phan bi~t c) De phU'crng trlnh (1) c6 hai nghi~m x va x th1 11'~0-2m+10~0m~5 Thea d!nh If Vi-etta c6 (*) x1 + x2 = _ m- {X1X2 1,o diem Ta c6 x; +x; +x x 2 2 -m- V~y = (x1 + x ) - x 1x = 1,o diem =88-m+3=16m=-5 (thea man) m = -5 Ia gia tri can tim a) Ta c6 ~ = [- ( 2a -1) J -4 [- (4a + 3)J= 4a - 4a + 1+ 16a + 12 2 = 4a + 12a + 13 = ( 2a + ) + > 0, V a V~y 1,o diem phU'ang trlnh da cho luon c6 hai nghi~m phan bi~t v&i moi a l x1 + X2 = - b = 2a -1 Cau b) Theo djnh li Vi-etta c6 _ c _a x 1x 4a-3 a 1,o diem Ta c6 ( x + X2 ) + x 1x = ( 2a -1)- 4a- =- V~y h~ thlic lien h~ giCPa hai nghi~m khong phv thu()c vao gia tri cua a Ia ( x + x ) + x 1x = - PHAN TRAC NGHieM (3,0 f>IEM) I 1-C I 2-A I 3-D I 4-A 5-A LU~N (7,0 f>IEM) PHAN cAu BIEU fliEM NQIDUNG a) Ta c6 bang gia tri I Y=x-x I -2 -1 -4 -1 -1 -4 o,5 diem 8o thi ham soy= -x2 Ia parabol di qua cac diem O(O;O); A(1;-1); B( -1;-1); C(2;-4); D(-2;-4) y -2 -1 o,5 diem - b) PhU'crng trinh hoanh d¢ giao di~m cua (P) va (d) Ia -x2 = 2mx-5 ~ x + 2mx-5 = 0,5 di~m Ta c6 ~' = m -1.( -5) = m + > 0, Vm V~y dU'ang thang (d) va parabol (P) luon cat t~i hai di~m phan bi~t v&i mQim Khi m = , ta c6 (d) : y = 4x- PhU'crng trinh hoanh dQ giao di~m cua (P) va (d) Ia -x2 = 4x- ~ x + 4x- = Ta c6 ~, = 2 - 1.(-5) = + = > PhU'crng trinh c6 hai nghi~m phan bi~t -2+~ = ~ y = -1; X2 = -2-~ = -5 ~ y = -25 V~ytQad¢haigiaodi~mcuadva (P) Ia M(1;-1) va N(-5;-25) X1 = 0,5 di~m a) Khi m = -5, ta dU'Q'C phU'crng trinh x + 4x- = Ta c6 a+ b + c = 1+ + ( -5) = nen phU'crng trinh c6 hai nghi~m phan bi~t Ia c -5 x1 = 1; x2 =- =- = -5 a V~y v&i m = -5 phU'crng trinh da cho c6 hai nghi~m 0,5 di~m 1,5 di~m x = va x =- b) Ta c6 ~'=2 -m= 4- m 8~ phU'crng trinh c6 nghi~m kep thi ~, = ~ 4- m = ~ m = Cau2 V~y m = Ia gia tri can tim 0,5 di~m c) 8~ phU'crng trinh ( 1) c6 hai nghi~m x1 va X2 th i ~' ~0 ~ 4-m~ ~ m :5:4 Thea dinh If Vi-etta c6 Ta c6 X +X2 { =-4 x; + x; = 10 ~ (x + x 2) - 1,0 di~m x1x2 =m 2x1x2 = 10 ~ ( -4 ) -2m = 10 ~ 16 -2m = 10 ~ m = (th6a man m :s;; ) V~y m = Ia gia tri can tim a) Ta c6 ~' = (m+5) -(6m -30) = m2 +10m +25 - 6m +30 = m +4m +55 2 =m +4m+4+51=(m+2) +51>0, Vm V~y phU'crng trinh da cho luon c6 hai ! nghi~m phan bi~t 1,0 di~m v&i mQi m x1 + X2 =- b = -2 (m + 5) Cau b) Thea d!nh li Vi-etta c6 _c_ x1x -6m-30 1,0 di~m a Ta c6 3(x1 +X2 )+x1x =-6(m+5)+6m-30 =-6m-30+6m-30=-60 V~y h~ thCrc lien h~ gill'a hai ( x1 + x2 ) + x1x2 = -60 nghi~m khong ph~:J thu¢c vao gia tri cua m Ia :: ~ DE SO PHAN TRAC NGHieM (2,5 £>1EM) I 1-B I I 2-B 3-B I 4-C 5-D PHAN TV' LUAN (7,5 £>1EM) cAu BIEU £>1EM NQIDUNG o,s diem Cau a) Ta c6 AE Ia dU'ang kinh cua dU'ang tron ( 0) nem JfCE =goo (g6c n¢i tiep chan nll'a dU'ang tron) b) Ta c6 7fBR = AEC (g6c n¢i tiep cung chan Ac ) Ma MBH vu6ng t(ili A nen BAH+ ABH =goo MCE vu6ng t(ili C nen 6AC + AEC =goo (1) (2) (3) TCP (1), (2) va (3) suy BAH= OAC A a) Xet tLI' giac ABHD c6 AHa= Cau AoB =goo (gia thiet) Ma hai dinh H, D ke cung nhln c(ilnh AB dU'&i m¢t g6c bang nen tLI' giac ABHD Ia tLI' giac n¢i tiep Tam Ia trung diem cua AB b) Trang MOB vuong t(ili D c6 OD Ia trung tuyen nen t(ili - - Suy OBD = ODB (hai g6c & day) Ma 6Bi5 = Hai5 (BE Ia phan giac cua ABC) 6i5B = Haf5 OD II BC Do d6, Ta cOng chi dU'Q'c: AH l BC, suy AH l 00 oo = OB => !1800 can c) Ta c6: BAH= HCE (cLing phv v&i g6c ABC) Ma BAR= Bi5R (tLI' giac ABHD noi ti~p dU'ang tron va g6c noi ti~p cung chan cung BH) Do d6, Bi5R = HcE (cung bang BAR) - - - - 1,0 di~m Ta IGii c6: BDH +HOE = 180° => HCE + HOE = 180° Khi d6 tLI' giac HDEC n9i ti~p Suy Fii5C = CER (hai g6c noi ti~p cung chan cung HC cua dU'ang tron ngOGii ti~p HDEC) B K F H c GQi M, N, E, F thea thLI' tv Ia cac ti~p di~m cua AB, AD, DC, CB v&i dU'ang tron (0) Thea tfnh ch§t hai ti~p tuy~n cat nhau, ta c6 AM=AN; ND=DE; CE=CF; BF=BM 0,5 di~m Suy AD+ BC =AN+ NO+ BF + FC =AM+ BM +DE+ CE = AB +DC Cau ~ AB + DC = AD+ BC = + 18 = 26 (em) ~ AB =DC =13(cm) Ke DH L BC tGii H va AK L BC tGii K Khi d6 ADHK Ia hlnh blnh hanh nen AK = DH va KH =AD= em Xet MBK va 11DCH c6 AB = DC va AK = DH Suy MBK = 11DCH (cGinh huy~n- CGinh g6c vu6ng) ~BK=CH BC-HK 18-8 = =S(cm) 2 Ap dvng djnh ly Py-ta-go vao !1DCH vuong t9i H, ta c6 Suyra HC= 0,5 di~m DH = DC -CH = 132 -5 = 144 => DH = 12cm V~y di~n tfch hlnh thang Ia SAaco = ;: ~ DES06 PHAN TRAC NGHieM (2,0 £>1t:M) I 1-C I 2-A I 3-C I 4-8 (AD+ BC).DH 26.12 =- 2 =156 ( cm 2) PHAN TV' LU~N (8,0 DIEM) BIEUDIEM NQIDUNG CAU The tfch IQ thLI' nhat: ~ = S.h = :rrR h ~ 3,14.15 20 ~ 14130 ( cm ) Cau The trch IQ thLI' hai: 3,0 diem V2 = S.h = :rrR h ~ 3,14.20 12 ~ 15072 ( cm ) Do ~ < V2 nen h~t nU'&c tl) IQ thLI' nhat sang IQ thLI' hai thl khong bi tran Di~n tfch hlnh tron (0; R) Ia S1 = :rrR Di~n tfch hlnh tron Di~n Cau (0; r) Ia S = :rrr 2 • • tfch hinh vanh khan Ia: r2 ) = :rr(17,5 - 7,5 ) Di~n tfch xung quanh hlnh n6n lEt S' = :rrrl = :rr 7, 5.30 S = S1 - S2 = :rr( R - 3,0 diem Vf}.y di~n tich can tlm Ia: S + S' = :rr( 17,52 -7,5 + 7,5.30) = 475:rr ( cm ) A B \12 \0 I \ 3/ C I I I \ I I \If ~ Cau 2,o diem E Goi A, B, c, o, E Ia cac diem nhU' hlnh ve £>~t h = AE , h-12 Taco: -h-= ~ h =48 (em) Vcoc 2.48- 3:rr3 ( 48 -12) = :rr ( 2.48-3 36 ) = 148:rr ( = 3:rr4 cm) • , n = 10.1000 GQl n I,a sof I"'an dong, ta co: :rr 148 Vf}.y An phai dong It nhat 22 lan PHAN TRA.C NGHieM (2,0 DIEM) 1-B 2-B 3-C 4-D 5-C ~ 21, PHAN TV' LU~N (8,0 DIEM) cAu BIE:U DIEM NQIDUNG 1) Ta c6: x = + /3 (thea man dieu ki~n xac dinh)~ -IX= + /3 Thay vao bieu thCPc A, ta dU'qc: A= o,5 diem =-2-= /3+1 1+~-2 J3-1 2) Ta c6: P =A + B ~P= = + JX _ 4fX+2 JX-2 x+1 xfX-2x+fX-2 2(x+ 1) -IX( fX -2) + 4-IX +2 ( jX- 2) (X+ 1) ( jX- 2) (X+ 1) ( jX- 2) (X+ 1) 1,o diem 3x-6fX = (-JX-2)(x+1) 3-IX =X+1 Cau 3-IX, Vq.y p = X+1 A VO'I X ~ T' , X =I= • 3) V&i X=O ta c6 P =0 V&i x =1= ta c6 P = fX+JX Ap d1,mg b!it d~ng thlfc CO-si cho hai s6 dU'ang JX va JX + Jx ~ 2~-JX Jx = Jx, ta c6: o,5 diem Do cl6 P ~~ oAu "=" xay va chi JX = Jx ~X= Vf!,y maxP=%, x=1 GQi X Ia so cay xanh moi dQi dU'Q'C theo dl)' ki~n (x E I~(' X< 300, cay) Thai gian d¢i hoan cong Cau vi~c theo k~ hoq.ch Ia 300 So cay thl)'c t~ d¢i dU'Q'C moi Ia X+ (cay) Thai gian d¢i hoan cong vi~c theo thl)'c t~ Ia (ngay) X 300 (ngay) X+ o,5 diem Thttc te d()i hoan cong vi~c sam han ngay, nen ta c6 phU'ang trlnh 0,5 di~m 300- 300 = x x+5 0,5 di~m Giai phl.rang trlnh ta dU'Q'C x = 25 (cay) V~y so cay d()i dtt kien tr6ng moi Ia 25 cay a) PhU'O'ng trlnh (1) c6 nghi~m X= -2, nen ta c6 m + 3) (-2) - m + = m = -15 V~y d~ phU'ang trlnh c6 nghi~m x = -2 thl m = ~15 (-2 ) 0,5 di~m - ( b) Ta c6: ~ = [- ( m + 3) r- ( -m + 5) = m +1Om -11 £>E3 phU'ang trlnh c6 nghi~m x1 ,x2 thl ~ [ Cau m < -11 m; (1) Thea dinh If Vi-et, ta c6: { X1 +X =ri7+3 ~ • ( 2) x1x2 m+5 0,5 di~m Thea bai rata c6: x/x2 + x1x/ = x1x2 ( X1 + x2 ) = (3) Thay ( 2) vao ( 3), ta dU'Q'C: (-m+5)(m+3)=7 ¢>[::~ m = m = Ia gia tri can tlm Ket hqp v&i ( 1) ta dU'Q'c V~y Cau a) Ta c6 K doi xll'ng v&i H qua BC nen KB = HB va KG= HC Do d6 ~BKC = !1BHC( c.c.c) ~ BKC = BHC Ta c6 BHC = FHE (doi din h) (1) (2) M~t khac t(r giac AEHF noi tiep ( AFR = 7fER = 90°) nen BAG +i=HE = 1aoo (3) 1,0 di~m TCJ' (1),(2) va (3) ta c6 8AC+8KC=180° Do d6 t(f giac ACKB n()i ti~p b) Ta c6 t(f giac BFEC n()i ti~p ( BFC = BEG = ME=Aca (cungbuv&i L~i (4) L~i goo XAB = Aca (5) 1,o diem XAB =ME===; xy II FE c6 OA _L xy nen OA c) Ta c6 ABA'= nen Bi=E) Ke ti~p tuy~n xy qua A (nhU' hinh ve) Khi d6 TCJ' ( 4) va ( 5) suy goo) _L FE hay A A' _L FE (vi AA' Ia dU'ang kfnh) hay A' B _LAB c6 CH _LAB===; A' BliGH 1,o diem TU'ang tV' ta c6 A' C II BH Do d6 t(f giac BHCA' Ia hinh binh hanh Ma /Ia trung diem cua BC nen dU'ang cheo th(f hai HA' phai qua / Hay ba diem H, I, A' thang hang va /Ia trung diem cua HA' d) Ta c6 0/la dU'ang trung binh cua MHA' nen OJ II AH - =HAG ===; 0/G (so le trong) va 0/ = -AH Mat khac G Ia tam cua MBC nen G/ =_!_GA o,5 diem Do d6 /1/GO(/)MGH(c.g.c) ===; TGO = AGR ===; H,G,O thang hang va OG = _!_HG Hai tam giac AGO va AGH c6 chung dU'ang cao va HG = 20G nen SMHG k = 2SMOG' ~ DE SO PHA.N TRAC NGHieM (2,5 DIEM) I 1-A I 2-B I 3-D 4-A 5-C PHA.N TV' LU~N (7,5 DIEM) cAu 1) a= + 4J3 (thoa man di~u ki~n xac din h) ===; Cau BIEU DIEM NQIDUNG Thay vao bieu th(fc P ta dU'Q'C: P= +4J3 _g 2+J3 -3 = 4-13-2 J3 -1 =5+-13 J8 = + J3 o,5 diem 2) Ta c6: 0= = = a-518 -3 + + 18-3 18 +3 a-9 ( 18 + 3) ( 18- 3) a - 518 - + a-9 + a-9 1,0 di~m a-9 a a-9 V~y Q =_a_, v&i a 0, a =t a-9 a a-9 3) Ta c6: P.Q = c va-3 a-9 = a 18-3· =J8+3+ Jag a -3 = ( J8 -3)+ Jag-3 + Ap d1,mg b§t d&ng thLPc Cauchy cho hai s6 dlPang J8- va Jag a -3 (do a >g), 0,5 cli~m ta c6: (Ja -3)+ Jag-3 ~2~(Ja -3) Jag-3 =6 o§u b&ng xay J8 -3 =Jag a -3 J8 -3 = 3(a >g) a= 36 Do cl6 A + = 12 V~y minA=12~a=36 GQi x Ia s6 chi tiet may ma to san xuat dl! cljnh lam mot theo ke ho(ilch (x E l~r*, x < 420, chi tiet may) Thai gian hoan cong vi~c theo dl! clinh: 420 (ngay) 0,5 di~m X s6 chi tiet may thi!C te to san xuat lam dU'Q'C mot Ia: X+ (chi tiet may) 420 ( ' ) th eo th l)'C tf e: - X+ Thi!C te to san xuat hoan cong vi~c sam han nen ta c6 phU'O'ng • hoan ' th'an h cong Th 0'' 1• g1an A Cau • A v1~c 0,5 cli~m trlnh 420 - 420 = X X+2 Giai phU'ang trlnh ta clU'Q'C x = 28 (chi tiet may) V~y s6 chi tiet may to san xuat dl! cljnh lam mot theo ke ho(ilch Ia 0,5 cli~m 28 chi tiet may EJ6p 6n a) PhU'ang tr1nh hoanh de) giao diem cua d va (P) Ia x = 2mx- 2m + x - 2mx +2m-3 = o,5 diem Ta c6 L1' = (-m) -(2m-3)= (m -1) + > 0, Vm V~y phU'ang tr1nh luon c6 hai nghi~m phan bi~t hay d luon cat (P) t9i hai diem bi~t phan b) Thea h~ th(fc Vi-et, ta c6 Cau =2m • Khi d6: =2m-3 X1 +X2 {X X 2 Y1 + Y2 < g X12 + x/ < g (x1 + x2 ) -2x1x1 -g < (2m ) 4m 2 - - (2m - 3) - g < o,5 diem 4m - < AD.8C = 8D.MC ( 1) 80 8 Lai c6: A = M va 7fi3M = l5BC nen M8M oo L).08C => A8.CD = AM.8D(2) 80 8C TCP (1) va (2) suy dieu phai chCPng h GQi I Ia diem chfnh giefa cung l&n 8C ta c6 81 = Cl Ap dl)ng djnh ly tren ta c6: 18.CD + /C.80 = ID.8C =>CD+ 80 = C.ID 18 Dau II= II s C 2R 18 khong doi xay /0 = 2R ~ 10 Ia dU'ang kfnh ~ D Ia diem chfnh giefa cung nh6 8C (vi I Ia diem chfnh giefa cung 8C l&n) o,s diem NHA XUAT BAN a~l HQC QUOC GIA HA NQI 16 Hang Chu6i - Hai Ba TrU'ng - Ha N()i Bi~n Quan thoQi: Bien t~p: (024) 39714896; ly xu~t ban: (024) 39728806; T6ng bien t~p: (024) 39715011; Fax: (024) 39729436 Chju trach nhi~m xu§t ban: Giam doc- T6ng bien t~p: TS PH~M THI TRAM Bien t~p chuyen nganh: BUI TRUNG HI~U, PHAM TH! OANH, HOANG LE THU HII~N, BANG TH! PHU'ONG ANH Bien t~p xu~t ban: PHAN HAl NHU' sera bai: PHfTH! KHANH VAN Ch~ ban: NGUY~N TH! NGQC HA Trinh bay bia: NGUY~N TH! HOANG DII;U Boi tac lien k~t: CQNG TV CO PHAN CCGROUP TOAN CAU 8ja chi: So 10 DU'ang Quang Ham, PhU'ang Quan Hoa, Qu~n CĐu Gi~y, TP Ha NÂi Ji! , I SACH LIEN KET I IIIII Maso: 1L-241PT2019 In 000 cuon kh6 22x27 5cm ,tQi cong ty CP in va ThU'O'ng mQi Quae Duy Bia chi: So ngach 130/1, ngo 130 Doc Ngii', P VTnh Phuc, Qu~n Ba Blnh, Tp Ha N9i So xac nh~n 8KXB: 4801-2019/CXBIPH/05-339/BHQGHN, 22/11/2019 Quy~t djnh xu~t ban so: 716 LK-TN/Qf)- NXB BHQGHN, 05/12/2019 In xong va n¢p IU'u chi~u nam 2019 ... 2x - yx- y = {2x- y = {y = 2x- { 2xy + 2x - 2xy - 3y = 2x - 3y = 2x - 3y = {y = 2x- {y = 2x- y = 2x - { 2x-3(2x-3)=1 2x-6x +9= 1 -4 x =-8 {y {y = 2x - =1 ¢::> X =2 ¢::> X =2" ... y&i nh?u 2x- y = -5 {4x- y = -1 { 3x+2y= 12 9x+6y=36 ¢::> 2x- 3y = { 13x = 26 -o ¢::> {2x- y = -5 13x = 26 {2x- 3y = -5 {2. 2- 3y = -5 X =2 X =2 ¢::> ¢::> {3y = X =2 {y = X =2 BU''&c Ket... 5y = 21 ~ B 12 ~ ~ E TRIN Ill ~ THAI AN Ill Narndu() ''9. 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