Thiết kế bài giảng đại số 10 nâng cao (tập 1) phần 2

ChirONq ill PHlTdNG TRINH VA HE PHl/dNG TRINH P h a n EOrolllirO VAIXr D E CUA C H I / O R T G L NOIDUNG Ndi dung chfnh cua chuong gom: Phuong trinh bae nha't : Giai va bien luan phuong trinh bae nhat cd chiia tham so, phuong trinh quy vl phuong trinh bae nhit - Phuang trinh bae hai : Giai va bien luan phuong trinh bae hai, dinh If Vi-et, mdt so phuong trinh quy vl phuang trinh bae hai He phuong trinh bae nhit va he phuong trinh bae hai II MUC Tl£U Kien thiirc Hieu khai niem phuong trinh, phuong trinh tuong duong, phuang trinh qua; bia't dugc cac phep bien doi tuong duong va phep bia'n doi cho phuong trinh qua Nim viing cdng thiic va cac phuong phap giai phuong trinh bae nha't, phuong trinh bae hai mdt in va he phuong trinh bae nhat, bae hai in Hieu y nghia hinh hgc ciia cac nghiem ciia phuong trinh va he phuong trinh bae nha't va bae hai KT nang Bia't each giai va bien luan : + Phuong trinh bae nha't va bae hai mdt in, + Phuang trinh dang |ax + b| = [ex + d| va phuong trinh chiia in d miu, 235 + Phuang trinh trung phuang + He hai phuong trinh bae nha't in (bing dinh thiic cip hai) • Bia't each giai (khdng bien luan): + He ba phuang trinh bae nhit in, + He phuang trinh bae hai Bia't giai mdt so bai toan vl tuang giao giiia dd thi cua hai ham sd bae khdng qua Thai - HS cd tfnh cin than, kidn tri va khoa hgc tim giao cua hai d6 thi HS tha'y dugc quan he mat thiet gifla toan hgc va ddi sdng, toan hoc xua't hiendo nhu ciu tur ddi sdng 236 P h a n a CAC B A I SOAJV §1 Dai cvtdng ve phtfofng trinh (tiet 1, 2) L MUC TifiU Kie'n thurc Giiip HS : Hiiu khai niem phuang trinh, tap xac dinh (diiu kien xac dinh) va tap nghiem ciia phuong trinh - Hia'u khai niem phudng trinh tuong duong va cac phep bien doi tuong duong KT nang Biet each thii xem mdt so cho trudc cd phai la nghiem ciia phuang trinh da cho hay khdng Bia't sii dung cac phep bia'n ddi tuong duong thudng diing Thai Ren luyen tfnh nghiam tiic khoa hgc II CHUAN BI CUA GV VA HS Chuan bi ciia GV : Chuin bi bai ki cac kia'n thiic ma HS da hgc d ldp de dat cau hdi Chuin bi mdt sd hinh ve SGK; pha'n mau, Chuan hi ciia HS : Cin dn lai mdt so kie'n thiic vl ham sd da hgc d ldp PHAN P H I T H I LUONG Bai chia lam tia't: Tie't thut nhdt: tic ddu de'n he't phdn 3; Tiet thvc hai: ta phdn den het phdn bdi tap 237 IV TIEN TRINH DAY HOC A Bai cii c a u hdi Hay cho bia't nghiem ciia phuong trinh 2x - = x + c a u hdi Cac sd nao sau day la nghiem cua phuong trinh x^ + x = V2x, 1;2;-1;0 B Bai mdi HOATDONGI Khai niem phuang trinh mot an Dinh nghTa Cho hai ham sd y = f(x) va y = g(x) cd tap xac dinh lin lugt la % va 2)g Dat 3) = 3)f n 2)^ Menh dl chiia bia'n f(x) = g(x) dugc ggi la mdt phuang trinh mdt in; X ggi la dn 50'(hay dn) va 3) ggi la tap xdc dinh cua phuong trinh Sd XQ e S) dugc ggi la mdt nghiem ciia phuang trinh f(x) = g(x) ne'u f(Xo) = g(Xo) la menh dl diing GV: Thuc hien thao tdc ndy tron^ phut Hoat ddng cua GV Hoat ddng ciia HS c a u hdi Cy'.fi y tra Idi cau hdi Hay neu mdt vf du vl phuang Chang ban Vx +1 = x + trinh mot in Ggi y tra Idi cau hdi c a u hdi Tap xac dinh ciia phuang trinh Hay nau tap xac dinh cua la [1; + 00) phuang trinh vira nau 238 Ggi y tra Idi cau hdi Cau hdi Hay chi mdt nghiem ciia Chang ban x = la nghiem phuang trinh Chu y 1) De thuan tien thuc hanh, ta khdng cin via't rd tap xac dinh ii) cua phuong trinh ma chi cin neu diiu kien de x e 3) Diiu kien dd ggi la diiu kien xac dinh ciia phuang trinh, ggi tit la diiu kien ciia phuang trinh Nhu vay, diiu kien ciia phuong trinh bao gom cac diiu kien de gia tri cu.a f(x) va g(x) ciing dugc xac dinh va cac diiu kien khac ciia in (ne'u cd yau cau) (theo quy udc vl tap xac dinh ciia ham sd cho bdi bieu thiic) GV: Thuc hien thao tdc ndy phdt Hoat ddng ciia HS Hoat ddng ciia GV c a u hdi Ggi y tra Idi cau hdi Hay ndu mdt thuat ngii khac Diiu kien xac^nh ciia phuong trinh ve tap xac dinh cua phuang hoac diiu kien cua phuang trinh trinh c a u hdi Ggi y tra Idi cau hdi Hay neu md'i quan he giiia tap nghiam va tap xac dinh cua Tap nghiam la tap cua tap xac dinh cua phuang trinh phuang trinh Vi du a) Diiu kien xac dinh ciia phuong trinh V x ^ - x ^ + = la x^ - 2x^ + > b) Khi tim nghiem nguyan cua j)huong trinh = Vx ta hieu diiu X kian ciia phuang trinh l a x e Z,X7tOvaj[:>0 (hay x nguyan duong) 239 Chu y 1) Khi giai mdt phuang trinh (tiic la tim tap nghiem ciia phuang trinh), nhilu ta chi cin, hoac chi cd the tfnh gia tri gin dung cua nghiam (vdi chinh xac nao dd) Gia tri dd ggi la nghiem gin dung cua phuang trinh Ching ban, bing may tfnh bd tiii, ta tfnh nghiem gin diing (chfnh xac da'n hang phin nghin) ciia phuong trinh x = la x « 1,913 2) Nghiem ciia phuang trinh f(x) = g(x) chfnh la hoanh cac giao diem cua thi hai ham sd y = f(x) va y = g(x) GV: Chi neu rdt nhanh cdc nhdn xet ciia chd y tren HOATDONG2 Phuong trinh tuong duang Ta da bia't : hai phuang trinh tuang duang na'u chiing cd cung mot tap nghiem Neu phuang trinh fi(x) = gi(x) tuong duong vdi phuang trinh f2(x) = g2(x) thi ta via't: fi(x) = gi(x) ^ f2(x) = g2(x) GV: Nhdn mqnh : - Hai phuang trinh ma tap nghiam ciia phuang trinh bang tap nghiem ciia phuong trinh thi tuong duong vdi - Hai phuong trinh cung vd nghiem thi tuong duong GV: Hudng ddn HS thuc Men|H1| vd thuc Men thao tdc ndy phdt Hoat ddng cua GV Hoat ddng cua HS c a u hdi Ggi y tra Idi cau hdi Tim tap nghiem cua phuang Tap nghiem ciia phuofng trinh la: trinh : S={1} Vx-l=2Vl-x 240 c a u hdi Ggi y tra Idi cau hdi Tim tap nghiem cua phuang Tap nghiam ciia phuong trinh la trinh : {1} X - = c a u hdi Ggi y tra Idi cau hdi Hai phuang trinh cd Hai phuang trinh tuang duang, tuang duang khdng? vi sao? vi chiing cd ciing tap nghiem Cau hdi Ggi y tra Idi cau hdi Tim tap nghiem cua phuang Tap nghiem cua phuong trinh la trinh X + Vx-2 = +Vx-2 Ggi y tra Idi cau hdi c a u hdi Tap nghiam cua phuang trinh la Tim tap nghiem ciia phuang {1} trinh X - = c a u hdi Ggi y tra Idi cau hdi Hai phuang trinh cd Hai phuang trinh khdng tuang tuang duang khdng? vi sao? duang, vi chiing khdng cd ciing tap nghiem c a u hdi Ggi y tra Idi cau hdi Tim tap nghiem ciia phuang Tap nghiam ciia phuang trinh la trinh {-i;i} 1x1 = Cau hdi Ggi y tra Idi cau hdi Tim tap nghiem ciia phuang Tap nghiem ciia phuong trinh la trinh {1} x = le-TKBGBAISOIONC-TI 241 c a u hdi Ggi y tra Idi cau hdi Hai phuang trinh cd Hai phuang trinh khdng tuang tuang duang khdng? vi sao? duang, vi chiing khdng cd ciing tap nghiem GV: Neu cdc nhdn xet sau : Khi mudn nha'n manh hai phuong trinh cd ciing tap xac dinh ^ (hay co ciing diiu kien xac dinh ma ta ciing kf hieu la 9^) va tuong duong vdi nhau, ta ndi : - Hai phuong trinh la tuong duong vdi tren ID, hoac - Vdi dieu kien if*, hai phuong trinh la tuong duong vdi Chang ban: Ta ndi, vdi x > 0, hai phuong trinh x = va x = la tuong duong vdi Trong cac phep bia'n doi phuong trinh, dang chii y nhat la cac phep bien doi khdng lam thay doi tap nghiem ciia phuang trinh Ta ggi chiing la cac phep bia'n doi tuong duang Phep Men ddi tuang duang Men mot phuang trinh thdnh phuang trinh tuang duang vdi nd Dudi day la dinh If vl cac phep bien doi tuong duong thudng diing Dinh Ii Cho phuang trinh f(x) = g(x) cd tap xdc dinh iP; y = h(x) la mot hdm sd xdc dinh tren iD (h(x) cd the Id mot hang sd) Khi dd tren y\ phuang trinh dd cho tuang duang vdi mdi phuang trinh sau : \)f(x) + h(x) ^g(x) + h(x); 2)f(x) h(x) = g(x) h(x) ne'u h(x) ^ vdi mgi A: e 2) GV: Hudng ddn HS chicng minh nhanh dinh li Tit dinh If tran, ta dl thay : hai quy tic bien doi phuong trinh da hgc d ldp dudi (quy tic chuyen ve va quy tic nhan vdi mdt sd khac 0) la nhiing phep bien doi tuong duong 242 GV: Hudng ddn HS thuc hien\H2\ vd thuc hien thao tdc ndy S phut Hoat ddng ciia HS Hoat ddng cua GV Cau hdi Ggi y tra Idi cau hdi O cau a), sau chuyen ve' Cd cd dugc phuang trinh tuang Theo dinh If tren duang hay khdng? Ggi y tra Idi cau hdi Sau luge bd ta dugc phuang b) Cho phuang trinh trinh 3x = x^ Phuang trinh cd 3x + V x - = x^ + V x - hai nghiam x = va x = nhung Luge bd V x - d ca hai ve X = khdng phai la nghidm ciia ciia phuang trinh thi dugc phuang trinh ban dau phuang trinh tuang duang Hai phuang trinh khdng tuang duang cau hdi HOAT DONG Phuong trinh he qua GV: Neu vi du de ddt vdn de ve phuang trinh he qua Xet phuong trinh : Vx = - X (1) Binh phuong hai ve, ta dugc phuong trinh mdi : X = - 4x + x^ (2) Tap nghiem ciia (1) la Sj = {1}, cua (2) la S2 ={ 1; 4} Hai phuong trinh (1) va (2) khdng tuong duong Tuy nhian ta thay S2 Sp Trong trudng hop ta ggi (2) la phuong trinh he qua ciia phuong trinh (1) Tong quat, fi(x) = gi(x) goi la phuang trinh he qua ciia phuong trinh f(x) = g(x) neu tap nghiem cua nd chita tap nghiem cua phuong trinh f(x) = g(x) Khi dd ta viet f(x) = g(x)=^fi(x) = g,(x) 243 GV: Thuc hien thao tdc ndy phiu Hoat ddng cua GV Hoat ddng cua HS c a u hdi Ggi y tra Idi cau hdi Hay ndu vf du ve hai phuang x = l = > x ^ = l trinh he qua: c a u hdi Ggi y tra Idi cau hdi Hay chi nghiem ngoai lai Nghiem ngoai lai cua phugng trinh (nghiem ngoai lai ciia phuang la-1." trinh la nhiing nghiem ciia phuang trinh he qua ma khdng phai la nghiem cua phuang trinh ban diu) Tii dinh nghia nay, ta suy : Neu hai phuang trinh tuang duang thi mdi phuang trinh deu la he qua ciia phuang trinh lai Trong vf du 2, gia tri X = la nghiem cua (2) nhung khdng la nghiem ciia (1) Nan X = la nghiem ngoai lai cua phuang trinh (1) GV: Huang ddn HS thuc hien |H3| vd thuc hien thao tdc ndy phiit Hoat ddng ciia GV Hoat ddng ciia HS c a u hdi Ggi y tra Idi cau hdi Khing dinh sau day diing hay Diing vi hai phuong trinh tuong sai? duong a) V x - = = > x - = 1? Ggi y tra Idi cau hdi c a u hdi Khang dinh sau day diing Diing, vi phuang trinh diu vd nghiem phuang trinh sau cd hay sai? nghiem x = , X x(x-l) b) -5^ ^ = l = > x = 1? x-1 Trong cac phep bie'n doi chi cho phuang trinh he qua, dang chu y la phep bia'n doi dugc nau dinh If sau 244 IL C Au HOI TRAC NGHI$M TU LUAN cau Giai va bien luan he phuong trinh sau: \ax + y = fx + y = [x-2ay = [ax-2y = l Cau Cho ham sd y = (2m - l ) x + 4x + m - a) Xac dinh m de ham sd dong bie'n trdn E b) Hay xac dinh m de ham sd cd gia ldn nha't cau Giai phuang trinh |x-l| +|x-2| = Hudng ddn vd ddp dn: I CAU HOI TRAC NGHIEM K H A C H QUAN cau (1 d) Chgn (b) Cau (1 d) Chgn (c) cau (Id) Chgn (a) cau (I d) Chgn (b) IL CAu HOI TRAC NGHIEM TU LU^N cau 1, (2 d, mdi cau d) a) D = -2a -I i^O, phuang trinh cd nghiem nhit b) D = - - a Nlu a = - phuang trinh vd nghiem Neu a ^ - phuong trinh cd nghiem nha't cau (2 d, mdi cau dilm) a) m >—; b)m < — cau (2diem) x = va x = 395 PHU LUC Ngudi ta bit diu phai nghi den doi mdi phuang phap day hgc d Viet Nam thdi dai vi tfnh hoa cau hdi dat la ta phai bit diu tir dau? Chic chin cau tra Idi cua nd la: Phai nghian ciiu va phai hgc Ta phai nghian ciiu de iing dung nhiing cai ma the gidi da cd vao thuc tiln Viet Nam Ta phai hgc vi neu khdng ta ching lam gi dugc Ddi vdi mdt giao vidn thdi hien dai viec diu tian la phai co trinh tin hgc co ban Dd'i vdi giao vian Toan diiu dd lai cin thiet va cap bach ban rat nhilu Sau day tdi xin gidi thieu mdt vai iing dung phin mim Maple qua trinh giang day dai so 10 A GlOl THIEU \ t PHAN MEM MAPLE Phang mIm cho ta tdc tfnh toan cue ki td't, md phdng hinh hgc mdt each true quan, giai phuong trinh va he phuang trinh thuan tien va mdt sd ling dung khac nira Dac biet la phan mIm cd giao dien vdi window ra't td't De gidi thieu vl nd cac ban, dgc tren cac webside ciia cac diin dan toan hgc B MOT SO tlNG DUNG vA VI DU I Tinh toan va tinh toan vdi cac sd gan dung Sau vao Maple, ta tfnh cac gia tri cua cac bieu thiic so vdi cac kf hieu sau: Phep cdng : Dau +; Phep nhdn : Dau *; Phep trie : Dau ; Phep chia : Dau /; Phep ndng len luy thita : 396 Dau ^ Phep tinh lay can bdc hai: sqrt Ngoai ta ciing dung cac dau ngoac (, {, ), },[,],••• Vi du T' K linh 1999^-2007' ^ , , ta lam nhu sau : 2005^+2008.2001 > (1999^2-2007^2)/(2005^3+2008*2001); -32048 8064168133 Sau nhap xong cac sd lieu an "Enter" may se cho ta kit qua Vi du 2: Tfnh gia tri ciia bieu thiic chinh xac din mudi chii sd thap phan Vi7+Vi3-2Vn 6V23 Ta lam nhu sau: > (sqrt(17)+sqrt(13)-sqrt(ll))/(6*sqrt(23)); — (Vi7+Vl3-Vn)V23 138 > e v a l f (%) ; 1533287151 Chii y rang lenh > e v a l f (%) ; Cd nghia la tfnh gia tri cua bieu thiic trudc dd de'n 10 chii sd thap phan Nlu ta cin tfnh trudc dd den sd chir sd thap phan bat ki, ta cd the lam nhu sau de tfnh gia tri bieu thiic tren din 10 chir sd (sau da nhap xong bieu thiic) >evalf((sqrt(17)+sqrt(13)-qrt(ll))/(6*sqrt(23)),5); 15333 Nhu vay neu chuang I, viec tfnh toan vdi sd gin diing se vd ciing thuan Igi nlu ta sii dung phin mIm 397 Vidu 3: 17 99 Trong hai sd— va — sd nao gin V2 ban (Bai tap dai sd 10 nang cao) 17 70 12 Tuy vao dau ciia bilu thiic ma ta danh gia kit qua Viec cd the sir dung may tinh bd tiii, song khdng dan gian vi nd cd gia tri tuyet dd'i De giai bai toan la chi cin thuc hien phep tinh V^- Ta thuc hien vdi Maple nhu sau: > (abs(sqrt(2)-17/12))-(abs(sqrt(2)-99/70)) ; 420 99 r Ta tha'y gia tri duong, do — gin v2 ban Vidu 4: Hay danh gia sai sd tuyet dd'i ciia so Pi ( TI ) vdi so 355 113 Gidi: Ta cd sai sd tuyet dd'i A = % - • 355 113 Sii dung phin mIm Maple bing cii phap sau ta cd: >abs(Pi-355/113) ; -Tt + 355 113 >evalf(abs(Pi-355/113),8); 10"^ Nhu vay, neu lay chinh xac den chii sd thap phan thi ta tha'y A < 0,2.10"^ 398 Ta cd the giai rat nhilu vf du, (trong sach giao khoa, sach bai tap 10) bing Maple ra't td't II Ham sd va gia tri ciia ham sd, thi cua ham so Ta hoan toan cd the tfnh dugc tfnh dugc gia tri ciia ham sd tai cac bien sd / Ddu tien chung ta dinh nghia hdm so, chdng hqn hdmf(x) = x^ + — > f:=x->x^2+x+l/3; /• /:=x^x+x +— > f (1) ; 7_ >f (sqrt(2)) ; + 42 >f (l-sqrt(3) ) ; (1-V3)2+ V3 >evalf(%); 137180910 > f (Pi) ; Tt +Tt+- >evalf(%); 13.34453039 > f (1/2) ; \3_ 12 399 >f (sqrt(3)-sqrt(2)) ; (V3-V2)2+V3-V2+-!3 >evalf(%); 7521910943 Ngudi ta cd the dung lenh unapply de chuyen mdt bieu thiJCc ve mot hdm sd, chdng hqn > g : = u n a p p l y ( x ' ^ + s q r t (2) * x + l - s q r t (2) ,x) ; g := X -> x^ + V2x +1 - V2 va tir dd ngudi ta cd the xac dinh dugc cac diem cin thiet ciia dd thi ham so, chang ban •) v Gia tri cue tieu: Tai x = gia tri cue tiau la >g(-sqrt(2)/2) ; i-V5 > evalf(%) ; 9142135620 Ta xac dinh dugc gia tri xap xi ciia nd Giao diem vdi true tung >g(0) ; I-V2 > e v a l f (%) ; 414213562 Hdm tdng khuc De xay dung ham tu"ng khiic ta diing lenh: Piecewise 400 Ching han >f :=piecewise(x restart; >with(plots) ; Warning, the name changecoords has been redefined 401 [animate^ animate danimatecurv^ arrow, changecoord?complexplofcomplexplot3(!l conformal conformaUdcontourplol contourplot3dcoordplo( coordplot3d cyliriderplo(densityplo(di.with(plottools); Warning, the name arrow has been redefined [arc, arrow, circle, cone, cuboi4 curve, cut in, cutout, cylinder, disk, dodecahedron ellipse, ellipticArc, hemisphere hexahedrori homothety hyperbolq icosahedror; line, oclahedroripieslice, point, polygon, project, rectangle, reflect, rotate, scale, semitorus, sphere, stellate, tetrahedron torus, transform, translate, vrml] Sau dd ta ve thi ham so bing cac cii phap sau: Vedd thi hdm thdng thudng , Ta ve thi ham sd y = f(x) bing cii phap Plot(f(x),x=a b,y=c d,titie= 'abed'); Ching ban Ve dd thi ham sd y = 2x Vl3 >plot(2*x-sqrt(13), 402 x=-10 10); Vdi thi ham sd trdn, ta khdng dua c, d vao thi chuong trinh tu ddng xac dinh miln gia tri y > p l o t ( * x ^ - * x + , x = - 4,y=2 ) ; 10 8i \i y \6 \ I—'—'—'—'—r -2 -1 —I—1—1—1—1—I—1—I—I—1—I '^ Ta cd the ve hai thi ciia hai ham so trdn ciing mdt he true >plot([2*x^2-3*x+5,2*x+l],x=-2 4,Y=2 10,color=[red,blue]); 403 Su van ddng cua dd thi Khi ham so cd chiia tham so, thi thi cua nd thay doi theo tham sd Khi tham sd biln thidn thi thi thay doi, ta ndi thi bien thian theo tham sd De lam, viec diu tian ta ciing nhap chuang trinh Sau dd, ta lam sach bd nhd bing lenh: Restart Sau dd nap gdi lenh chiic nang chuyan dung de ve thi bing hai lenh lian tia'p: With(plots) With (plotttools), nhu tran Ching ban > animate ( x ' ' + * t * x - t - l , x = - 404 , t = - l , color=red) ; 40- 30- 20- 10- • • 1' • -4 '— -2 X Sau dd, ta danh dau vao thi, in chudt ban trai, vao nhirng not diiu khien ma ta thfch, sau dd a'n Play IV Giai phuang trinh va he phuang trinh Budc Tao lap phuang trinh bing each gan cho phuang trinh dd mdt kf hieu nao dd, chang han Eqn: =, PTl :=, Budc Dung lanh solve de giai phuong trinh Solve(eqn); Vi du > q n : = x'^3+2*x-3=0; eqn :=x^ + x - = > solve(eqn) ; 1 1 2 i,_i+l/Vn,-^-^/Vn 2 Ta tha'y phuong trinh cd nghiem thuc x = va hai nghiem phiic - - - - / V n va - - + - / V n 2 2 405 Vi du >PT:=sqrt(x-1)=3*x-5; PT:=4^ = 3x-5 >solve(PT) ; Vi du >ab:= sqrt (5-x)+sqrt (x+3) =3; ab:=45-x +4x + =3 > solve(ab) ; l+lV7 1-^V7 2 Vi du >solve((abs(x+abs(x+2))^2-1)^2=9.{x}); {x = 0},{xeqn3:= x+7*y-4*z=22; eqn3 :=x + 7y - z = 22 Budc Diing lenh solve dl giai he > solve({eqnl,eqn2,eqn3},{x,y,z}) ; f -334 -36 -2511 19 406 19 19 MUC LUC • • Trang Chuang L MENH Di - TAP HOP §1 Menh dl va menh dl chiia bia'n §2 Ap dung menh dl vao suy luan toan hoc Luyen tap §3 Tap hgp va eac phep toan tren tap hgp Luyen tap §4 So gin diing va sai sd cau hdi va bai tap dn tap chuong I Chuang n HAM SO BAC NHAT VA BAC HAI §1 Dai cuong vl ham sd Luyen tap §2 Ham sd bae nhit Luyen tap §3 Ham sd bae hai Luyen tap Cau hdi va bai tap dn tap chuong II Chuang /// PHUONG TRJNH VA HE PHL/ONG TRiNH § Dai cuong vl phuong trinh §2 Phuong trinh bae nhat va bae hai mdt in Luyen tap §3 Mdt so phuang trinh quy vl phuong trinh bae nhat hoac bae hai Luyen tap §4 He phuong trinh bae nhat nhilu in Luyen tap §5 Mdt so vf du vl he phuang trinh bae hai hai in Cau hdi va bai tap dn tap chuang III ON TAP HOC Ki I PHU LUC 32 45 53 75 90 108 124 125 151 163 176 187 198 212 235 237 258 283 291 300 310 330 337 346 369 396 407 Chiu trdch nhiem xudt bdn : Giam ddc: DINH NGOC BAO Tong bian tap: LE A Chiu trdch nhiem ndi dung vd bdn quyen CONG TY TNHH SACK GIAO DUC HAI ANH Biin tap ndi dung : NGUVfeN TIEN TRUNG My thuat - Vi tinh : THAI SON SON LAM Trinh bdy bia : THANH HUYEN THIJ^T KE BAI GIANG DAI SO 10 - NANG CAO, T^P MOT In 1000 cudn, kho 17 x 24cm, tai Cdng ti co phin in Phiic Yan Sd dang kf KHXB : 219.2006/CXB/80 - 25/DHSP 28/3/06 In xong va nop luu chieu thang 11 nam 2006 [...]... chgn ket qua diing 12 Cho hai phuong trinh : |x|= 1 (1) va x^ - 3x + 2 = 0 (2) (a) (1) la ha qua ciia (2) ; (b) (2) la he qua cua 1); (c)(l)« (2) ; (d) ca ba ket luan tren diu sai Hay chgn ket qua diing 13 Cho hai phuong trinh x+l + ^ i = = - 2 Vx + 1 (1) va x^ + 2x + 5 = 0 Hay chgn ket qua diing trong cac khing dinh dudi day: (a) (1) la he qua cua (2) ; (b) (2) la he qua cua 1); (c)(l)c^ (2) ; (d) ca ba ket... - 4 X + 1 = 0 Hay giai phuang trinh nay bing 2 each : tfnh A hoac A' Ggi y tra Idi cau h d i l 2x = - 2 0 va S > 0 thi 0 < x^ < X2 (hai nghiem duong) - Na'u P > 0 va S < 0 thi Xj < X2 < 0 (hai nghiem am) Vi du 4 Phuang trinh ( l - V 2 ) x ^ - 2 ( l + V2)x + V 2 = 0 cd : a = l - V 2 < 0 va c = V 2 > 0 =>... trinh : x^-2x + m - l = 0 Hay xac dinh cac he so a va b Hoat ddng cua HS Ggi y tra Idi cau hoi 1 a = 1; b = -2; c = m - 1 26 1 cau hdi 2 Ggi y tra Idi cau hdi 2 Hay giai va bien luan theo m A' = l - m + l = 2 - m phuang trinh nay Neu A' < 0 ... vax2, Xl + X2 bing (a )20 -V5; (b)V20-V5 ; (c )20 +V5; (d)V20 + V5 ~ 2 Hudng ddn : Xj + X2 = (xj + X2) - 2xi.X2 22 Phuang trinh 2x - 3x - = cd nghiem Xi va X2 ma Xi + X2 bing (a)- (b)-; (0- 12 (d)-... Xi va X2 thoa man fxi+X2 =3 (a) ' ' (b) M [xi.X2=l [xi.X2 =2 f x i +X2 = f x i +X2 = (c) ^ ^ • [xi.X2=-l 28 0 fxi +XT =3 ^^A [xi.X2=0 20 21 Phuong trinh x^ - 3x + V5 = cd nghiem |xi - X21 bing... l)(x-V2) X- 41 V2+1, = (x-V2)r(V2+l)x-V2 Hien nhian phuong trinh cd hai nghiem Ta cd Xi+ X2 = va X1X2 =-15 27 2 Bai 10 Hudng ddn Hien nhian phuang trinh cd hai nghiem Ta cd XY+ X2 = va XjX2 = -15 a)