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

- 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.. Kien thiirc Hieu khai niem phuong trinh, phuong trinh tu

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ChirONq ill PHlTdNG TRINH VA HE PHl/dNG TRINH

P h a n 1 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

1 Kien thiirc

Hieu khai niem phuong trinh, phuong trinh tuong duong, phuang trinh

ha qua; bia't dugc cac phep bien doi tuong duong va phep bia'n doi cho phuong trinh ha 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 2 in

Hieu y nghia hinh hgc ciia cac nghiem ciia phuong trinh va he phuong trinh bae nha't va bae hai

2 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,

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+ Phuang trinh trung phuang

+ He hai phuong trinh bae nha't 2 in (bing dinh thiic cip hai)

• Bia't each giai (khdng bien luan):

+ He ba phuang trinh bae nhit ha 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 2

3 Thai do

- HS cd tfnh cin than, kidn tri va khoa hgc khi 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

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Ren luyen tfnh nghiam tiic khoa hgc

II CHUAN BI CUA GV VA HS

Bai nay chia lam 2 tia't:

Tie't thut nhdt: tic ddu de'n he't phdn 3;

Tiet thvc hai: ta phdn 4 den het phdn bdi tap

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IV TIEN TRINH DAY HOC

1 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^ 3 phut

Hoat ddng cua GV

c a u hdi 1

Hay neu mdt vf du v l phuang

trinh mot in

c a u hdi 2

Hay nau tap xac dinh cua

phuang trinh vira nau

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Cau hdi 3

Hay chi ra mdt nghiem ciia

phuang trinh

Ggi y tra Idi cau hdi 3

Chang ban x = 1 la nghiem

Chu y 1

1) De thuan tien trong 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 trong 3 phdt

Hoat ddng ciia GV

c a u hdi 1

Hay ndu mdt thuat ngii khac

ve tap xac dinh cua phuang

trinh

c a u hdi 2

Hay neu md'i quan he giiia tap

nghiam va tap xac dinh cua

phuang trinh

Hoat ddng ciia HS Ggi y tra Idi cau hdi 1

Diiu kien xac^nh ciia phuong trinh hoac diiu kien cua phuang trinh

Tap nghiam la tap con cua tap xac dinh cua phuang trinh

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Chu y 2

1) Khi giai mdt phuang trinh (tiic la tim tap nghiem ciia phuang trinh), nhilu

khi ta chi cin, hoac chi cd the tfnh gia tri gin dung cua nghiam (vdi do 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 = 7 la x « 1,913

2) Nghiem ciia phuang trinh f(x) = g(x) chfnh la hoanh do cac giao diem cua do thi hai ham sd y = f(x) va y = g(x)

GV: Chi neu rdt nhanh cdc nhdn xet ciia chd y tren

HOATDONG2

2 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 nay bang tap nghiem

ciia phuong trinh kia thi tuong duong vdi nhau

- Hai phuong trinh cung vd nghiem thi tuong duong

GV: Hudng ddn HS thuc Men|H1| vd thuc Men thao tdc ndy trong 5 phdt

Tap nghiem ciia phuofng trinh nay la:

S = { 1 }

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Hai phuang trinh nay cd

tuang duang khdng? vi sao?

Ggi y t r a Idi cau hdi 3 Hai phuang trinh nay tuang duang,

vi chiing cd ciing tap nghiem

Ggi y t r a Idi cau hdi 4 Tap nghiem cua phuong trinh nay la 0

Ggi y t r a Idi cau hdi 5 Tap nghiam cua phuang trinh nay la {1}

Ggi y t r a Idi cau hdi 6 Hai phuang trinh nay khdng tuang duang, vi chiing khdng cd ciing tap nghiem

Ggi y t r a Idi cau hdi 7 Tap nghiam ciia phuang trinh nay la {-i;i}

Ggi y t r a Idi cau hdi 8 Tap nghiem ciia phuong trinh nay la {1}

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c a u hdi 9

Ggi y t r a Idi cau hdi 9

Hai phuang trinh nay khdng tuang 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 nhau tren ID, hoac

- Vdi dieu kien if*, hai phuong trinh la tuong duong vdi nhau

Chang ban: Ta ndi, vdi x > 0, hai phuong trinh x = 1 va x = 1 la tuong duong vdi nhau

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 1 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) ^ 0 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

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GV: Hudng ddn HS thuc hien\H2\ vd thuc hien thao tdc ndy trong S phut

Hoat ddng cua GV

Cau hdi 1

O cau a), sau khi chuyen ve'

cd dugc phuang trinh tuang

ciia phuang trinh thi dugc

phuang trinh tuang duang

Cd

Theo dinh If tren

Sau khi luge bd ta dugc phuang trinh 3x = x^ Phuang trinh nay cd hai nghiam x = 0 va x = 3 nhung

X = 0 khdng phai la nghidm ciia phuang trinh ban dau

Hai phuang trinh nay khdng tuang duang

HOAT DONG 3

Phuong trinh he qua

GV: Neu vi du 2 de ddt vdn de ve phuang trinh he qua

Xet phuong trinh :

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)

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GV: Thuc hien thao tdc ndy trong 2 phiu

Hay chi ra nghiem ngoai lai

(nghiem ngoai lai ciia phuang

trinh la nhiing nghiem ciia

phuang trinh he qua ma

khdng phai la nghiem cua

phuang trinh ban diu)

Hoat ddng cua HS Ggi y t r a Idi cau hdi 1

x = l = > x ^ = l

1

Ggi y t r a Idi cau hdi 2 Nghiem ngoai lai cua phugng trinh la-1."

Tii dinh nghia nay, ta suy ra : Neu hai phuang trinh tuang duang thi mdi phuang trinh deu la he qua ciia phuang trinh con lai

Trong vf du 2, gia tri X = 4 la nghiem cua (2) nhung khdng la nghiem ciia (1) Nan X = 4 la nghiem ngoai lai cua phuang trinh (1)

GV: Huang ddn HS thuc hien | H 3 | vd thuc hien thao tdc ndy trong 2 phiit

Ggi y t r a Idi cau hdi 2 Diing, vi phuang trinh diu vd nghiem con phuang trinh sau cd nghiem x = 1

Trong cac phep bie'n doi chi cho phuang trinh he qua, dang chu y la phep bia'n doi dugc nau trong dinh If sau

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Djnh II 2 Khi binh phuong hai ve cua mdt phuong trinh, ta dugc phuong

trinh he qua cua phuong trinh da cho Ndi each khac :

Hay chiing to day la phep

bia'n doi he qua

Hoat ddng cua HS Ggi y tra Idi cau hdi 1

Vx +1 = 2x +1 ^ X +1 = f 2x +1 j ^

Phuang trinh dau chi cd nghiem

X = 0; nhung phuang trinh sau cd

3 hai nghiam x = Ova x =

4

Chii y

1) Cd the chiing minh ring na'u hai va' ciia mdt phuang trinh ludn cung ddu thi khi binh phuong hai ve' cua nd, ta dugc phuong trinh tuong duong

2) Na'u phep bien doi mdt phuong trinh din den phuong trinh he qua thi

sau khi giai phuong trinh he qua, ta phai thit Iqi eae nghiem tim dugc vao

phuong trinh da cho de phat hien va loai bd nghiem ngoai lai

GV: Thuc Men thao tdc ndy trong 2 phiit

c a u hdi 1

Hay dat diiu kien cho x de khi

binh phuong hai va' ciia phuong

Diiu kien x > —

2

Binh phuang hai ve' ta dugc :

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3

4x^ + 3x = 0 o X = 0 va x = - - Kiem

4 tra dieu kien ta duoc x = 0 la nghiem

GV: Neu vd hitdng ddn HS gidi vi du 3 trong SGK hoac cd the thay baitg vi du

khdc tuang tie

HOATDONG 4

4 Phuong trinh nhieu an

Trong thuc te, ta con gap cac phuong trinh cd nhilu ban mot in Dd la

cac phuong trinh dang F = G, trong dd F va G la cac bieu thiic cua nhilu bie'n

Neu phuong trinh hai in x va y trd thanh menh dl diing khi x = Xy va y = y,, (vdi

Xo va y„ la cac sd) thi ta ggi cap sd (x^{, y,,) la mdt nghiem ciia phuong trinh

Ching ban, cap sd (1; 0) la mdt nghiem ciia phuong trinh (3)

Khai niem nghiem ciia phuong trinh ba in, bd'n an, ciing dugc hieu

tuong tu Ching han bd ba sd (1; 1; 1) la mdt nghiem cua phuong trinh (4) Ddi

vdi phuong trinh nhieu in, cac khai niem : tap xac dinh (diiu kien xac dinh),

tap nghiem, phuong trinh tuong duong, phuong trinh he qua, ciing tuong tu

nhu ddi vdi phuong trinh mdt in

GV: Thitc hien thao tdc ndy trong 5'

X + y = 5

(0;5),(1;4),

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5 Phuong trinh chiia tham sd

Chung ta con xet ca nhiing phuong trinh, trong do ngoai in x con cd

nhiing chii khac Cac chii nay dugc xem la nhiing sd cho trudc va dugc ggi la

Khi m = 0, hay chi ra phuang

trinh va tap nghiem

c a u hdi 2

Hay chi ra mdt nghiem

phuang trinh khi m ^ 0

manh y dd khi giai phuang trinh chiia tham so, ta thudng ndi la gidi vd Men ludn phuang trinh

247

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TOM TAT BAI HOC

1 Cho hai ham sd y = f(x) va y = g(x) cd tap xac dinh lin lugt la iDf va 9")^

Daty) = ii)f n ^ \

Menh dl chiia bien f(x) = g(x) dugc ggi la mdt phuong trinh mdt in;

X ggi la in sd (hay in) va 9) ggi la tap xac dinh ciia phuang trinh

Sd XQ e ^ ggi la mdt nghiem ciia phuong trinh f(x) = g(x) na'u f(Xo) =

g(Xo) la menh di dung

2 Hai phuong trinh tuong duong ne'u chiing cd cung mdt tap nghiem Neu phuong trinh fi(x) = gi(x) tuong duong vdi phuang trinh f2(x) = g2(x) thi

ta via't:

fi(x) = gi(x)c^ f2(x) = g2(x)

3 Dinh li 1

Cho phuong trinh f(x) = g(x) cd tap xac dinh iT); y = h(x) la mdt ham sd xac

dinh trdn 9) (h(x) cd the la mdt hing so) Khi do trdn ^, phuong trinh da cho

tuong duang vdi mdi phuang trinh sau :

l)f(x) + h(x) =g(x) + h(x);

2) f(x) h(x) = g(x) h(x) neu h(x) 9^ 0 vdi mgi x e ^J)

4 fj(x) = gi(x) ggi la phuong trinh he qua ciia phuang trinh f(x) = g(x) neu tap nghiem ciia nd chiia tap nghiem ciia phuong trinh f(x) = g(x) Khi do

ta viet

f(x) = g(x)=>fi(x) = gi(x)

5 Dinhli2

Khi binh phuong hai vi ciia mdt phuong trinh, ta dugc phuong trinh he

qua cua phuong trinh da cho Ndi each khac :

f(x) = g ( x ) ^ [ f ( x ) f = [g(x)]^

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HUdNG D A N T R A LCJI C A U HOI vA BAI T A P SGK

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vi khdng thoa Vay phuong trinh vd nghiem

x # l

Vdi diiu kien x ;^ 1, ta cd :

phuang trinh cd nghiem x =2 Trd Idi:

lb) Vdi dieu kien x ^t 2, ta cd :

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x + - i _ ^ 2 x _ 3 ^ x ^ - 2 x + l = 2 x - 3

x - 2 x - 2 c^ X = 2 (loai do diiu kien x T^ 2) Vay phuong trinh vd nghiem

Chu y : Binh phuong hai v6 ta chi thu dugc phuong trinh he qua nan

phai thii lai de ket luan nghiem

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V x - l = x - 3

=» X - 1 = (x - 3)^ => x^ - 7x + 10 = 0

=> X = 2 hoac X = 5

Thii lai, gia tri x = 2 khdng thoa man

Vay phuong trinh cd nghiem x = 5

c) Ta cd :

2| X - 1 1 = X + 2

^ 4(x -1)^ = (X + 2f ^ 3x^ -12x = 0

=> X = 0 hoac X = 4

Thii lai thay ca hai diu nghiem diing

Vay phuang trinh cd hai nghiem x = 0 va x = 4 d) Ta cd :

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( c ) | x | = l ; ( d ) x ^ + V x = 1+Vx

4 Cho phuong trinh

x^ + x + V x + l = 0 (1) Hay diin diing - sai vao cac ket qua sau day:

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Hay chgn diing - sai trong cac khang dinh sau

D D u n g DDiing DDiing

(a) phuong trinh (1) vd nghiem Vm;

(b) phuong trinh (1) cd 3 nghiem Vm;

(c) phuong trinh (1) cd 2 nghiem la x = - 1 va x = 2 - m;

(d) ca ba ket luan tren diu sai

Cho phuong trinh

1

X +

x + 1 = 1 (a) phuong trinh (1) cd 2 nghiem la x = - 1 va x = 0;

(b) phuong trinh (1) cd 2 nghiem la x = 2 va x = 0;

(c) phuong trinh (1) cd 2 nghiem la x = 1 va x = 2;

(d) ca ba ket luan tran diu sai

Hay chgn khing dinh diing

7 (d)

3 (c) (c)D, (c) Sai;

(d)D (d) Diing

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BAITAPTUGIAI

8 Cho phuong trinh

x^ + 2005x + VxTI = 1 (1) Hay chgn ket luan dung trong cac ket luan sau

(a) X = 1 va X = 5 la nghiem cua (1);

( b ) ( l ) o ( x - l ) ( x - 2 ) ;

( c ) ( l ) o ( x - l ) ( x + 2);

( d ) ( l ) o ( x + l ) ( x - 2 )

Hay chgn ket qua diing

10 Cho phuong trinh

x + Vx = 0 ' (1) ( a ) ( l ) o x = O v a x = 1;

( b ) ( l ) c ^ x = 0;

(c)(1) < » x = 1 vax = 2;

( d ) ( l ) c ^ x = 0 v a x = - 2

11 Cho phuong trinh

2 / - 1 1

X + vx + v;^

x-3 Vx-2

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(a) Diiu kien ciia phuong trinh la : x > 0, x 9^ 3;

(b) Diiu kien ciia phuong trinh la : x > 2;

(c) Diiu kien eiia phuong trinh la : x > 2, x ?!: 3;

(d) Diiu kien ciia phuong trinh la : x > 0

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

13 Cho hai phuong trinh

x + l + ^ i = = - 2 (1)

Vx + 1

va

x^ + 2x + 5 = 0 (2) Hay chgn ket qua diing trong cac khing dinh dudi day:

(a) (1) la he qua cua (2);

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(d) ca ba ka't luan tran diu sai

16 Cho phuang trinh : x"^ + 3x^ + 2 = 0

(a) Phuang trinh cd 2 nghiem x = - 1 va x = - 2 ; (b) Phuong trinh cd 4 nghiem x = ±1 va x = ±2; (c) Phuang trinh vd nghiem;

(d) ca ba ket luan tran diu sai

Hay chgn ka't qua diing

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§2 Phufdng trinh bae nhat va bae hai mot an

Bia't each bien luan sd giao diem cua mdt dudng thing va mdt parabol

va kiem nghiem lai bing do thi

Biet ap dung dinh If Vi-et de xet dau ciia mdt phuong trinh bae hai va bien luan sd nghiem cua mdt phuong trinh triing phuang

• Can dn lai mdt so kien thiic vl ham sd da hgc

Can dn lai phin phuong trinh da hgc d ldp 9 va bai 1

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i n P H A N PHOI THCII LI/ONG

Bai nay 2 tiet :

Tie't thic nhdt: tu: ddu de'n he't phdn 3;

Tie't thic hai: td phdn 4 de'n he't phdn bdi tap

IV TIEN TRINH DAY HOC

Phuang trinh bae hai (mot in) la phuong trinh ed dang

ax + bx + c = 0 (a, bvacla ba sd da cho vdi a ^ 0);

2

Ta cd: A = b - 4ac ggi la Me

thdc thu ggn eiia phuong trinh bae hai

Ta cd: A = b^ - 4ac ggi la Met thitc A' = b'^ - ac (vdi b = 2b') ggi la Met

259

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Trong bai nay, chiing ta se nghidn ciiu each giai va bien luan cac phuong trinh cd dang phuong trinh bae nha't hoac phuang trinh bae hai va cd chiia tham sd

GV: Thuc Men thao tdc ndy trong 5 phiu

Hay giai phuang trinh nay

bing 2 each : tfnh A hoac A'

Cach 2 A' =

X l

-X2 =

: 4 :24

2) a = 0 va b 9^ 0 : Phuong trinh vd nghiem

3) a = 0 va b = 0 : Phuong trinh nghiem diing vdi mgi x e R

GV: Thuc hien thao tdc ndy trong 3 phut

2

a = m - l ; b = - m + l

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Hay xac dinh cac he sd a va b

cau hdi 2

Hay giai va bien luan theo m

phuang trinh nay

Khi a ^ 0 <» m # ±1, phuong trinh cd nghiam duy nhit x =

m + 1 Khi m = 1 : ta thiy a = 0, b = 0 phuong trinh cd vd sd nghiem

Khi m = - 1 : a = 0, b = 2 7^ 0 phuang trinh vd nghiem

GV : Hudng ddn HS ldm vi du 1 GV hudng ddn HS gidi bang cdch ddt vdn de

HI Phuong trinh da cho tuong duong vdi phuang trinh nao?

H2 Hay chia cac trudng hgp va bien luan

H3 Ka't luan nghiem

HOAT DONG 2

2 Giai va bien luan phuang trinh dang ax^ + bx + c = 0

1) a = 0 : Trd vl giai va bien luan phuong trinh bx + c = 0

2) a ?!: 0 : Neu A < 0, phuong trinh vd nghiem;

Na'u A = 0, phuong trinh cd mdt nghiem (kep) x = Na'u A > 0, phuong trinh cd hai nghiem (phan biet)

-GV: Thitc Men thao tdc ndy trong 3'

a = 1; b = -2; c = m - 1

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cau hdi 2

Hay giai va bien luan theo m

phuang trinh nay

A' = l - m + l = 2 - m Neu A' < 0 <:5' m > 2 phuong trinh vd nghiem

Na'u m = 2 phuang trinh cd nghiem kep X - 1

Na'u m < 2 phuong trinh cd 2 nghiem phan biet

GV: Hudng ddn HS ldm vi du 2 GV hudng ddn HS gidi bang cdch ddt van de

HI Phuong trinh tran da la phuong trinh bae hai chua?

H2 Chia cac trudng hgp va bien luan

H3 Tfnh A neu dd la phuong trinh bae hai

H4 Ka't luan nghiem

Chd y

1 •> ,

Trong vf du 2, khi m = 4 ta cd Xj = X2 = — nan ta cd tha chi nau ba trudng hgp : m > 4 ; 0 ; ^ m < 4 v a m = 0 (trudng hgp 0 ^^ m < 4, phuong trinh co nghiem la Xj va X2)

GV: Hudng ddn HS thuc hien\\\^ trong 3 phut

c a u hdr 1

Phuang trinh da cho cd the vd

nghiem dugc hay khdng?

c a u hdi 2

Phuang trinh ludn cd 2

nghiem phan biet cd diing

khdng?

Khdng vi ta thiy x = 1 la nghiem

Tiir phuong trinh

X - mx + 2= 0

tathay : (1 - m ) x = - 2 Neu m = 1, phuong trinh nay vd nghiem, phuong trinh da cho cd |

Trang 29

nghiem duy nhat x = 1

Na'u m 7^ 1, phuong trinh nay cd

2 2 nghiam x = va neu = 1

m - 1 m - 1

<:^ m = 3 phuong trinh nay cd nghiem

X = 1, phuong trinh da cho cd nghiem kep X = 1

Vdi m 9^ 3, phuang trinh da cho cd hai nghiem phan biet

GV : Hudng ddn HS ldm vi du 3

HI Hay dua phuong trinh vl dang f(x) = a

H2 Hay khao sat va ve do thi ham sd y = f(x)

H3 Bien luan so nghiem phuong trinh bing do thi

GV: Hudng ddn HS thuc Men | H 2 | GV treo hoac vehinh 3.1 len bdng vd thitc

Men thao tdc ndy trong 3 phiu

c a u hdi 1

Dua vao hinh 3.1, tim cac gia

tri ciia a de phuang trinh (3)

cd nghiem duang

c a u hdi 2

Trong trudng hgp dd, hay tim

nghiem duang ciia phuang

trinh

Phuong trinh (3) cd nghiem duong khi phuong trinh (4) cd nghiem duong

Dua vao hinh 3.1 ta tha'y a > 2 phuong trinh cd nghiem duong

Nghiem duang cua phuang trinh la nghiem ldn ciia phuang trinh (4)

Giai (4) ta dugc x = -1 +V2 + a

Chu y

Khi via't phuong trinh (3) dudi dang x^ + 3x + 2 = x + a, ta thay ka't qua

tran con cho bia't sd giao diem cua parabol y = x + 3x + 2 vdi dudng thing

y = X + a

263

Trang 30

HOATDONG3

3 tTng dung ciia dinh Ii Vi-et

6 ldp dudi, chiing ta da hgc dinh If Vi-et dd'i vdi phuong trinh bae hai va cac ling dung cua nd

Ta nhic lai hai iing dung quan trgng sau :

1) Na'u da thiic f(x) = ax + bx + c cd hai nghiem Xj va X2 thi nd cd th6' phan tfch thanh nhan tii

Da thiic da cho cd nghiem la :

x = 1, X = 6

f(x) = ( x - l ) ( x - 6 ) 2) Ne'u hai so cd tdng la S va tfch la P thi chiing la cac nghiem ciia phuang trinh x^ - Sx + P = 0

GV: Hudng ddn HS thuc hien\HZ\ vd thuc Men thao tdc ndy trong 5 phiit

cau hdi 1

Ne'u ggi chilu dai va chieu

rdng ciia hinh chir nhat la a va

b thi ta cd bieu thiic nao ?

c a u hdi 2

Hay lap phuang trinh cd hai

nghiem la a va b trong timg

trudng hgp

a + b = 20

a b = P

a) Vdi P = 99, phuang trinh la x^ 20x + 99 = 0 , X i = 9 , X 2 = l l

-Ta phai khoanh hinh chu nhat kfch thudc 9 cmxll cm

Trang 31

b) Vdi P = 100, phuong trinh la

Vay trong trudng hgp nay, khdng

cd hinh chir nhat nao thoa man yau

ciu di bai

Nhan xet dudi day suy ra true tiep tut dinh If Vi-et Nd cho phep ta nhan bia't da'u cac nghiem ciia mdt phuong trinh bae hai ma khdng cin tfnh cac nghiem dd

2

Cho phuang trinh bae hai ax + bx + c = 0 cd hai nghiem Xj va X2 vdi gia

b c thiat rang Xj < X2 Kf hiau S = — va P = — Khi dd :

a a

- Na'u P < 0 thi Xj < 0 < X2 (hai nghiem trai da'u)

- Na'u P > 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)

Trong vf du tran, ca hai ka't luan (phuong trinh cd hai nghiem, va hai

nghiem do trai da'u) diu dugc suy ra tii P < 0

265

Trang 32

Trudng hgp P > 0, ta phai tfnh A (hay A') de xem phuong trinh cd nghiem

hay khdng roi mdi tfnh S di xac dinh da'u cac nghiem

GV: Hudng ddn gidi vi du 5 Sau dd, GV hudng ddn HS thuc hien\H4\ (GV

thuc hien thao tdc ndy trong 5 phiit)

Cd ka't luan gi ve nghiem

phuang trinh a)

Cau hdi 3

Hay lam tuang tu dd'i vdi

phuang trinh b)

Hoat ddng cua HS Ggi y t r a Idi cau hdi 1

a va c trai dau

Ggi y t r a Idi cau hdi 2 phuang trinh a) cd hai nghiem trai da'u

Chgn (A)

Ggi y t r a Idi cau hdi 3 b) Phuong trinh

x^ - ( V 2 + V I ) x + Vd = 0 cd hai nghiem phan biet Xj va X2 vi:

Do do ta chgn phuong an (B)

Ta da bia't: ddi vdi phuong trinh triing phuong

ax^ + bx^ + c = 0, (4) neu dat y = x (y > 0) thi ta di den phuong trinh bae hai dd'i vdi y :

ay^ + by + c = 0 (5)

Trang 33

Do dd, mudn biet sd nghiem cua phuong trinh (4), ta chi cin biet sd

nghiem cua phuong trinh (5) va diu cua chiing

GV: Hudng ddn HS thitc Men H 5 va thuc Men thao tdc ndy trong 5 phiu

c a u hdi 1

a) Na'u phuang trinh (4) cd

nghiem thi phuang trinh (5)

chic chin cd nghiem

Dung hay sai?

c a u hdi 2

b) Na'u phuang trinh (5) cd

nghiem thi phuang trinh (4)

chic chin cd nghiem

Diing hay sai?

Diing

b) Sai, vi khi phuang trinh (5) chi cd nghiem am (hoac mdt nghiem kep am, hoac hai nghiem am phan biet) thi phuong trinh (4) vd nghiem

GV: Hudng ddn HS ldm vi du 6 trong SGK Ngodi ra GV cd the cho HS ldm

them vi du sau :

Vi du

Cho phuang trinh :

x"^ - 2(m + 3)x^ + 2m + 5 = 0 (6) Vdi gia tri nao cua m thi phuong trinh (6) cd bdn nghiem phan biet?

Gidi:

2

Dat y = X (y > 0), ta di dan phuong trinh :

y^ - 2(m + 3)y + 2m + 5 = 0 (7) Hien nhian, phuong trinh (6) cd bd'n nghiem phan biet khi va chi khi

phuang trinh (7) cd hai nghiem duong phan biet

Vi (7) cd hai nghiem y = 1 (do a + b + c = 0) va y = 2m + 5 nan diiu kien

de phuong trinh (6) cd bd'n nghiem phan biet la

0 < 2m + 5 ;>t 1 hay - 2,5 < m ^ - 2

26/

Trang 34

TOM T A T B A I HOC

1 Giai va bien luan phuong trinh dang ax + b = 0

1) a 9t 0 : Phuong trinh cd nghiam duy nhat x = — •

a 2) a = 0 va b ;^ 0 : Phuang trinh vd nghiem

3) a = 0 va b = 0 : Phuong trinh nghiem diing vdi mgi x e R

2 Giai va bien luan phuong trinh dang ax^ + bx + c = 0

1) a = 0 : Trd vl giai va bien luan phuong trinh bx + c = 0

2) a ?^ 0 : Na'u A < 0, phuong trinh vd nghiem;

Na'u A = 0, phuong trinh cd mdt nghidm (kep) x = ;

2a Neu A > 0, phuang trinh cd hai nghiem (phan biet)

3 VI dinh If Vi-et

2 ' 1) Na'u da thiic f(x) = ax + bx + c cd hai nghiem Xj va X2 thi nd cd the

phan tfch thanh nhan tu f(x) = a(x - Xi)(x - X2) (xem bai tap 9)

2) Na'u hai sd cd tong la S va tfch la P thi chiing la cac nghiem ciia

- Ne'u P < 0 thi Xj < 0 < X2 (hai nghiem trai da'u)

- Neu P > 0 va S > 0 thi 0 < Xj < X2 (hai nghiem duong)

- Neu P > 0 va S < 0 thi Xj < X2 < 0 (hai nghiem am)

Trang 35

Sai vi chua cd tap xac dinh

Ggi y tra Idi cau hoi 2

Sai, vi chua so sanh nghiem vdi tap tap xac dinh ciia phuang trinh

phuang trinh nay da la

phuang trinh bae hai chua

c a u hdi 2

Hay giai va bien luan phuang

trinh

Chua vi ha sd a chiia tham sd

2 m - 3

X =— (voi mot m) m^+1

Trd Idi cdc cdu edn Iqi

b) Vdi m = 1, phuong trinh nghiem diing vdi mgi x

269

Trang 36

Vdi m 9t 1, phuong trinh cd nghiem duy nhat x = m + 2

c) Vdi m ^ 2 va m ;t 3, phuong trinh vd nghiem

Vdi m = 2 hoac m = 3, phuong trinh nghiem diing vdi mgi x

Ggi y m(x - m + 3) = m(x - 2) + 6 <:5' Ox = m^ - 5x + 6

O Ox = (m - 2)(m - 3)

m

m - 2 d) Vdi m ^ 1 va m :?!: 2, phuong trinh cd nghiem x =

Vdi m = 1, phuong trinh nghieni diing vdi mgi x

Vdi m = 2, phuong trinh vd nghiem

Phuang trinh nay da la

phuang trinh bae hai chua?

Trd Idi cdc cdu con Iqi

b) Phuong trinh vd nghiem trong m

a = b = Ovac;tO

a ;^ 0 va A = b^ - 4ac < 0

Chua vi ha sd a chiia tham so

Goi y tra Idi cau hdi 2

Phuong trinh cd mdt nghidm trong mdi trudng hop sau :

a = 0 va b 9^ 0

a ;t 0 va A = b^ - 4ac = 0

5i trudng hgp sau :

Trang 37

Bai 8

GV: Hudng ddn giai cdu a)

Hoat ddng cua GV

cau hdi 1

Phuang trinh nay da la

phuang trinh bae hai chua?

Chua vi he sd a chira tham sd

Khi m = 1, phucmg trinh trd thanh 3x - 1 = 0, cd mdt nghiem x = —

Khi m :;<!: 1, ta cd phuang trinh bae

hai vdi biet sd

A = 9 + 4 ( m - l ) = 4m + 5

Vdi m > — - phuong trinh cd hai

4 -3±V4m + 5

Thay vao ta thay ngay Xj va

la cac nghiem cua phuong trinh

X2

271

Trang 38

hai nghiem Xj va X2

c a u hoi 2

Chiing td :

ax^ + bx + c = a(x - Xi)(x - X2)

Ggi y tra Idi cau hoi 2

Vi hai phuong trinh

2

ax + bx + c = 0 va a(x - Xi)(x - X2) = 0 tuong duong

Trd Idi cdc cdu con lai

Cdch khdc cdu a)

h c a) Ta cd cac ha thiic Xi + X2 = — va X1X2 =— Do dd :

Trang 39

Hudng ddn : Chii y ring phuang trinh bae hai tuong iing cd ac < 0 nan nd

cd hai nghiem trai da'u, suy ra phuong trinh da cho cd diing hai nghiem dd'i

nhau Tii do ta loai cac phuang an (A) va (C) Phuang an (D) ciing bi loai bing

each thii true tia'p

273

Trang 40

BO SUNG KIEN THtTC

De giai phuong trinh, ddi khi ta diing each dat in phu

Gia sii ta cin giai phuang trinh sau vdi diiu kien D :

F(x) = 0 (1) Neu F(x) cd the viet dugc dudi dang F(x) = f(u), trong dd u = g(x) thi

viec giai phuong trinh (1) cd the tien hanh nhu sau :

Budc 1 : Dat in phu

Trong phuang trinh (1), dat u = g(x) ta cd phuang trinh an la u (thudng

ggi la phuong trinh trung gian) :

f(u) = 0 (2)

Budc 2 : Giai phuong trinh (2)

- Neu (2) vd nghiem thi (1) vd nghiem

- Neu (2) cd cac nghiem la Ui, U2, thi vdi diiu kien D :

gfxj = ui (1) O gfxj = U2

Budc 3 Giai cac phuong trinh g(x) = Ui, g(x) = U2, vdi diiu kien D

Khi dd tap nghiem ciia phuong trinh (1) la tap tat ca cac nghiem thu dugc

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