Thiết kế bài giảng giải tích 12 (tập 2) phần 2

Ndi dung chinh cua chuang 4 : - So phiic : Dinh nghia ; hai sd phiic bdng nhau; bieu dien hinh hgc ciia so phiic; md dun ciia sd phiic T sd phiic lien hgp.. Cac phep toan ve sd phiic :

Ndi dung chinh cua chuang 4 :

- So phiic : Dinh nghia ; hai sd phiic bdng nhau; bieu dien hinh hgc ciia so phiic;

md dun ciia sd phiic T sd phiic lien hgp

Cac phep toan ve sd phiic : Phep cdng va phep trii; phep nhan cac sd phiic ; Tong va tfch hai sd phiic lien hgp ; phep chia hai sd phiic

Phuong trinh bac hai ddi vdi he sd thuc : Can bac hai ciia sd thuc am; phuong trinh bac hai ddi vdi he so thuc

n MUC TIEU

1 Kien thurc

Ndm dugc toan bd kien thiic co ban trong chuang da neu tren, cu the :

Ndm viing dinh nghia so phiic va cac phep toan ciia nd

• Hieu dugc mddun cua so phiic va bieu dien mdi sd phiic tren mat phang tga do Mdi quan he ciia hai so phiic lien hgp

2 KT nang

Van dung thanh thao cac phep toan

• Tim dugc mddun cua mdt so phiic

• Giai dugc phuang trinh bac hai cd nghiem phiic

3 Thai do

Tur giac, tich cue, dgc lap va chii dgng phat hien ciing nhu linh hdi kien thirc trong qua trinh hoat dgng

Cam nhan dugc su cdn thie't eiia dao ham trong viec khao sat ham sd

Cam nhan duoc thuc te' ciia toan hgc, nha't la ddi vdi dao ham

Phan 2

CAC B A I HOAX

§1 So' phu'c (tiet 1, 2, 3)

I MUC TIEU

1 Kien thurc

HS ndm dugc :

So i la gi? Y nghia ciia nd

• Dinh nghia so phiic

Hai so phiic bdng nhau khi nao?

Bieu dien hinh hgc sd phiic

Mddun ciia sd phiic

2 KT nang

HS tfnh mddun cua so phiic

• Tfnh thanh thao so phiic lien hgp cua mdt sd phiie

3 Thai do

- Tu giac, tich cue trong hgc tap

• Biet phan biet rd cac khai niem co ban va van dung trong tung trudng hgp cu the

- Tu duy cac van de cua toan hgc mdt each Idgic va he thdng

n CHUAN BI CUA GV VA HS

1 Chuan hi cua GV

• Chudn bi cac cau hdi ggi md

• Chudn bi phan mau, va mdt so dd diing khac

2 Chuan bj cua HS

m PHAN PHOI THCJI LUONG

Bai nay chia lam 3 tiet:

Tii't I : Tit ddu den hit muc 3

Tii't 2 : Tii'p theo din hit muc 5

Tiet 3 : Tii'p theo den hit muc 6 vd hudng ddn bdi tap

IV TIEN TRINH DAY H O C

HI Cd nhiing sd am nao khi binh phuang thi bdng 1

H2 Cd nhiing sd am nao khi binh phuang thi bdng - 1

H3 Phai chang cd mdt so khdng la so phiic ma khi binh phiuong bang - 1 ,

• GV neu khai niem' sd i:

Nghiem cua phuang trinh x +1 = 0 Idso i

2

Nhu vay : i = - 1

HOAT DONG 2

2 Djnh nghTa so phiic

• GV neu dinh nghia sd phiic:

Mdi bieu thdc dgng a + bi; a, b e R i = - 1 duac ggi la mot so phicc Ddi vdi sd phicc z = a + bi, ta noi a la phdn thuc, b la phdn do cua z Tap hgp cdc sdphdc ki hieu Id C

C = {a + 6i I a, 6 e R, i^ = - l )

H4 Hay neu vi du ve so phutc

H5 Sd thuc la trudng hgp rieng ciia sd phiic Diing hay sai?

• Thuc hien QL 1 trong 5'

GV ggi mdt vai HS tra Idi

Phdn thuc :,,,

Phdn ao :,,,

Sau dd ket luan

Ggi y tra Idi cau hdi 2

GV ggi mdt vai HS tra Idi

Phdn thuc :

Phan ao :

Sau dd ket luan

Ggi y tra Idi cau hdi 3

GV ggi mdt vai HS tra Idi

Phdn thuc :

Phdn ao :

Sau dd ket luan

H6 Phai chang ca phdn thuc va phdn ao cua mdt so phiic la mdt so thuc?

HOAT DONG 3

3 Hai sd phurc b^ng nhau

• GV neu dinh nghia :

Hai sdphitc Id bdng nhau neu phdn thuc vd phdn do cua chung tuang dng bdng nhau

a + bi = c + di<::>a = c v a b = d H7 Hay neu mdt so vi du ve hai so phiic bdng nhau

H8 Cho sd phiic : V2 + 3i Sd nao sau day bdng sd phiic tren

( a ) 2 - V 3 i ; ( a ) - 2 - V 3 i

4 + 2V3i

• Thuc hien vi du 2 trong 5' GV cd th^ thuc hien vi du khac

Hoat dgng ciia GV

Cau hdi 1

Mdi quan hd ciia x va y de hai

sd phiic dd bang nhau

Cau hdi 2

Tim X va y

Hoat dgng cua HS Ggi y tra Idi cau hdi 1

2x + l = x + 2 v a 3 y - 2 = 3/ + 4

Ggi y tra Idi cau hdi 2

HS giai he tren ta cd : x = 1 va y = 3 H9 Tim cac so thuc x va y, biet

(x + 1) + iSy - 2)1 = i-x + 2) + (2y + 4)i

HIO Tim cac so thuc x vay, biet

(-X + 1) + (2y - 1)1 - (x + 2) + (y + 4)1

• GV neu chii y :

• Mdi sd thuc a dugc coi la mdt sd phiic vdi phdn do bdng 0

a = a + Oi

Nhu vay, mdi so thuc cung la mot sophHtc Ta cd R c C

• Sd phiic 0 + bi dugc ggi la sd do va vie't dan gian la bi

bi = 0 + bi

Dac biet i = 0 + li

So / dugc ggi la dan vi do

H l l Hay chi ra phdn thuc va phdn do cua cac sd sau:

a) 7 ; b) -4i

Sd nao la sd thudn do?

• Thuc hien ^ ; 2 ti'ong 5'

1 ^

z = 1

2 2 Ggi y tra Idi cau hdi 2

GV ggi mdt vai HS tra Idi

z = yj2 + 5i

Ggi y tra Idi cau hdi 3 z= V 2 - 5 i

HOAT DONG 4

4 Bieu di^n hinh hgc ciia sd phirc

• GV neu dinh nghia :

Diem M(a ; b) trong mot he tog do vuong goc cua mat phdng duac ggi

la diem bieu diin sd phutc z = a + bi

• GV sir dung hinh 68 de dat cac cau hdi:

H12 Bieu dien cac so phiic sau tren mat phdng:

a ) l + 3i; b ) 2 + V3i;

HI3 Tim tap hgp cac sd phiic tren mat phang tga do chi cd phdn ao

HI4 Tim tap hgp cac sd phiic tren mat phang tga do chi cd phdn thuc

HI5 Hai sd phiic dugc bieu diln tren mat phang tga do cd dac diem gi ne'u:

a) Cd phdn thuc bang nhau nhung phdn ao ddi nhau

b) Cd phdn do bdng nhau nhung phdn thuc ddi nhau

c) Cd phan thuc va phdn do ddi nhau

• Thuc hien -^ 3 trong 5'

5 Md dun cua sd phiic

• GV neu dinh nghia :

Do ddi cua vecta OM duac ggi la mddun cua sd phicc z vd ki hieu la jzj

Vay \a + bi\ = Va^ + b^

• Thuc hien ^ 4 trong 5'

Va^ + &^ = 0 » a = & = 0

H16 Mdi sd phiic deu cd mdt mddun Diing hay sai?

H17 Hai sd phiic bdng nhau cd mddun bdng nhau Diing hay sai?

HI8 Hai so phirc cd mddun bdng nhau thi bdng nhau Diing hay sai?

Bieu dien hai sd :

2+ 3i va 2 - 3i tren mat phang

• GV neu dinh nghia :

Cho sd phicc z = a + bi Ta ggi a - bi la sd phiic liin hop cua z vd

ki hieu la 'z - a - bi

H19 Hai sd phiic lien hgp cd ciing mddun Dung hay sai

H20 Hay neu phan vi du ve hai sd phiic cd ciing mddun nhung khdng phai hai sd phiic lien hgp

• Thuc hien Ql 6 trong 5'

• GV neu va thuc hien vi du 4 GV cd the thay bdi vi du khac

H21 Tim sd phiic lien hgp ciia z = 13 -5i

H22 Tim sd phiic lien hgp cua z = -13 -5i

H23 Tim sd phiic lien hcrp ciia z = 13 +5i

H24 Tim sd phiic lien hgp ciia z = -13 + 5i

HOATTX:>NG 7

TOM TflT Bfll HQC

1 Mdi bieu thiic dang a + &/; a, 6 e R i = - 1 dugc ggi la mdt sd phurc

Ddi vdi so phiic z = a + bi, ta ndi a la ph^n thuc, b la phan ao ciia z

Tap hgp cac sd phiic kf hieu la C

C = {a + 6i | a , 6 e R, i^ = l }

-2 Hai sd phiic la bang nhau ne'u phdn thuc va phdn ao ciia chiing tuong ling bdng

nhau a + 6i = c + (ii<=>a = c v a 6 = (i

3 Dilm Mia ; 6) trong mdt he toa do vudng gdc cua mat phang dugc ggi la diem

Dien diing sai v^o chd trdng sau:

Cdu 1 Cho sd phiic z = 2 - 5i

(a)z=2 + 5i • (b)|z| = V29 •

f a - 2 (c)z = z ' «

Cdu 4 Cho hai sd phiic z = 2 - yi, z' = 5x + 3i; z = z' khi

2 (a) X = 2, y = - 3 ;

(c) V2T;

Trd Idi (b)

(d) V153

Cdu 9 Cho sd phiic z = vl2 - 3 i; |z| bang

: 12 - 3 i ; |z| bang

(b) Vl5 (d) VTsI

HOAT DONG 9

md^^ DflN Bfll TflP SflCH GIflO KMOfl

Bai 1 Hudng ddn z = a + bi thi phdn thuc a, phdn do b

GV cho HS len bang dien vao d trdng

Phdn thuc

2V2,

Phdnao

0

d)

Phdn thuc

0

Phdn ao -7

Bai 2 Hudng ddn Six dung cac tfnh chdt ciia hai sd phiic bang nhau

f3x-2 = x + l

cau a Giai he

2y + l = - ( y - 5 ) r>' -' 3 4

Ddpsd X = 0, y = 1

B^i 3 Hudng ddn Six dung cac tinh chat ciia sd phiic

cau a Thudc dudng thdng x = - 2

Ddp sd Hinh ve y^

-2 O

cau b Thudc dudng thang y = 3

Bai 4 Hudng ddn Six dung cac tfnh chdt ciia mddun sd phiic

Neu z = a + ib thi |z| = Va^ + b^

cau a Ddp sd I z I = 'ji-2f+iSf = 4l

Cau b Ddp sd \z\ = ^[(Sf^+i^ = V n

cau c Dap sd \ z \ = \]i-5) - 5

cau d Ddp sd I z_| = ^[iSf = V3

Bai 5 Hudng ddn Tinh ehdt ciia mddun so phiic tren mat phang tga do

Cau a Ddp sd La dudng trdn ban kfnh 1

cau b Dap sd La hinh trdn ban kfnh 1

y |

cau c Dap sd La phdn hinh gidi han bdi hai hinh trdn ban kinh 1 va 2

y*

caud

Ddp sd La dudng trdn ban kinh 1

Bai 6 Hudng ddn Dua vao dinh nghia sd phiic lidn hgp

cau a Ddp sd z =1 + iyf2

cau b Ddp sd z = -V2 - iy/s

cau a Ddp sd z = 5

Cau a Ddp sd z = -71

§2 C o n g tru* v a n h a n so' phuTc

(tiet 4, 5)

I MUC T l £ u

1 Kien thurc

HS ndm dugc :

Khai niem phep cdng, phep trii so phiie

- Dinh nghia phep cdng sd phiic

- Dinh nghia phep trir sd phiic

Phep nhan so phiic

2 KT nang

Van dung thanh thao cac phep toan cdng va trii so phiic

Ket hgp cac tinh chdt de thuc hien eac phep toan

So sanh vdi cac phep toan ciia sd Ihuc

3 Thai do

- Tu giac, tich cue trong hgc tap

Biei phan biet rd cac khai niem co ban va van dung trong tumg trudng hgp cu thi

- Tu duy cac vdn de ciia toan hgc mdt each Idgic va he thdng

n CHUAN BI CUA GV VA HS

1 Chuin bj cua GV

Chudn bi eac cau hdi ggi md

Chudn bi pha'n mau va mdt sd dd diing khac

2 Chuan bj cua HS

Cdn dn lai mdt sd kie'n thiic da hgc bai 1

m PHAN P H 6 I THCJI LUONG

Bai nay chia lam 2 tie't:

Tiit I : Tic ddu din hit phdn 1

Tie't 2 : Tii'p theo din hit phdn 2

IV TIEN TRINH DAY HOC

A OAT VAN OE

Cau hdi 1

Neu cac khai niem vl sd phiic :

- Dinh nghia sd phiic

- Sd phiic lien hgp

- Mddun cua so phiic

- Bieu diln hinh hgc eiia sd phiic

Cau hdi 2

Tim so phiic lien hgp cua cac sd sau:

a)z = 4 - 7 i ; b ) z = V 3 - 5 i Cau hdi 3

Tim mddun cia cac so phiic sau :

a)z = 4 - + 7 i ; b ) z = V 3 + 5 i

B BAI Mdl

HOAT DONG 1

1 Phep cdng va phep trur

• Thuc hien ^ 1 trong 5'

Cau hdi 3

Tfnh (-V3 + 5 i ) - ( 4 - 3 i )

Ggi y tra Idi cau hdi 3

(-VJ + 5i) - (4 - 3i) = (-V3 - 4) + 2i

• GV neu dinh nghia

Phep cong vd phep triJC hai sdphCcc duac thUc hien theo quy tac cong, trie da thdc

HI Tinh (7 + 5i) + (4 + 3i)

HS tu tinh

Ggi y tra Idi cau hdi 2

HS tu tinh

' GV neu tong quat:

(a + bi) + (c + di) = (a + c) + (b + d)i ;

(a + bi) - (c + di) = (a - c) + (b - d)i

H6 Tim so phiic lien hgp ciia z- (2 - 3i) - (5 + 4i)

H7 Tun sd phiic lien hgp ciia z- (2 - 3i) + (5 + 4i)

H8 Tim sd phiic lien hgp ciia z = (2 - 3i) - (5 - 4i)

H9 Tim sd phiic lien hgp cua z = (2 + 3i) - (5 + 4i)

(3 + 2i)(2 + 3i) = 6 + 13i + 6i2 = 13i

Ggi y tra Idi cau hdi 2

(3 + 2i)(2-3i) = 6 - 5 i - 6 i 2 = 1 2 - 5 i

• GV neu dinh nghia

Phep nhdn hai sdphdc duac thuc hien theo quy tdc nhdn da thicc roi

•2

thay i - -1 trong kit qua nhdn duac

- Thuc hien vi du 2 trong 5' GV cd the Idy vi du khac

HS tu tfnh

Ggi y tra Idi cau hdi 2

HS tu tfnh

HIO Tim sd phiic lidn hgp ciia z = (2 - 3i)(5 + 4i)

Hll Tim so phiic lien hgp ciia z = (2-3i)(5 + 4i)

HI2 Tim sd phiic hen hgp ciia z = (2 - 3i)(5 - 4i)

H13 Tim sd phiic lien hgp cua z- (2 + 3i)(5 + 4i)

• GV neu tong quat:

(a + bi)ic + d o = (ac - bd) + iad + bc)i

i = - 1 trong ket qua nhan dugc

(a + bi)ic + di) = iac - bd) + iad + bc)i

HOAT DONG 4

MQT SO CflU HOI TR^C NGHIEM KHflCH QUflN

Hdy dien dung sai vao 6 trdng sau:

Cdu 7 Cho z = 2 + 4i, z' = 5 - 3i

(a) z la so phiic lien hgp cua z'

Hdy chgn khdng dinh dung trong cdc cdu sau:

Cdu 5 Cho z = (3 + 2i) + (5 - i) So phiic lien hgp cua z la

z ' = i

(b) 1 (d)i

-HOAT DONG 5

- 3 3i

Bai 2 Hudng ddn Six dung dinh nghia phep cong, phep trir sd phiic

Cau a Ddp sd a + P = 3 + 2i, a - p = 3 - 2i ;

cau b Ddp sd a + P = l + 4i,a-p = l~Si ;

Cau c Ddp sd a + p = -2i, a - p = \2i

cau d Ddp sd a + p = I'd - 2i, a - p = ll + 2i

Bai 3 Hudng ddn Six dung cac tfnh chdt ciia phep nhan sd phiic vdi i^ = - 1 Cau a Ddp sd - 1 3 i

Bai 5 Ap dung cac hdng dang thiic

cau a Hudng ddn (2 + 30^ = 4 + 12i + (30^ - - 5 + 12i

cau b Hudng ddn Ta cd (2 + 30^ = 8 + 3.4.3i + 3.2(30^ + (30^

= 8 + 36/ - 54 - 27f = - 4 6 + 9 j

HOAT DONG 6

Bfll TflP BO SUNG

1 2 + 3i Bai 1 Chiing minh

Bai 2 Chiing minh

Bai 3 Chiing minh

Bai 5 Chiing minh z + z' = z + z'

Bai 6 Chiing minh z.z' - zz'

§3 Phep chia so phuTc

( t i e t 6, 7)

I M U C TIEU

1 Kien thiic

HS ndm dugc :

Nghich dao cua mdt sd phiic la gi ?

Phep chia hai sd phiic dugc thuc hien nhu the nao ?

Bai toan tong va tich ciia hai sd phiic lien hgp

2 KT nang

• Tim dugc nghich dao cua mdt sd phiic

- Thuc hien d.ugc phep chia hai sd phiic

3 Thai do

Tu giac, tfch cue trong hgc tap

Biet phan biet rd cac khai niem co ban va van dung trong timg trudng hgp cu the

Tu duy cac vdn d l ciia toan hgc mdt each Idgic va he thdng

n C H U A N BI CUA GV VA HS

1 Chuan bj cua GV

• Chuan bi cac cau hdi ggi md

• Chudn bi pha'n mau, va mdt sd dd diing khac

2 Chuan bj cua HS

Can dn lai mdt sd kien thiic da hgc d hai bai trudc

On tap kl bai 2

HI P H A N PHOI T H 6 I L U O N G

Bai nay chia lam 2 tiet :

Tii't 1 : Tit ddu din hit dinh nghia phep chia hai sdphdtc

Tie't 2 : Tii'p theo din hit

IV TlfiN TRINH DAY HOC

\ Tong va tich ciiia hai sd phurc lien hgp

• Thuc hien - ^ 1 trong 5'

Ta cd z = 2 - 3 i Ggi y tra Idi cau hdi 2

z + z = 4 Ggi y tra Idi cau hdi 3 z.z = 13

Ggi y tra Idi cau hdi 4

GV de HS tu nhan xet va kd't luan

HI Nhan xet vl z + z

H2 Nhan xet ve z.z

H3 Thuc nhien tiep cac phep tinh sau :

z + z ^

z.z =

• GV neu dinh nghia

Tong cua mot sd phicc vdi sd phvCc lien hap cua no tfdng hai Idn phdn thuc cda sdphdc do

Tich cua mot sd phicc vdi sd phicc lien hap cua no bdng binh phuang mddun cua sdphicc do

H4 Tong hai sd phiic lien hgp la mdt sd phiic hay sd thuc

H5 Nhan xet vl phdn thuc ciia so phiic tong dd

• GV neu nhan xet:

Vay tong vd tich cua hai sdphiCc lien h(yp Id mot sdthuc

GV cho HS tu dat va thuc hien eac phep tfnh sau de hinh thanh kl nang:

z z z + z z - z z z

HOAT DONG 2

2 Phep chia hai sd phurc

• GV neu dinh nghia :

Chia sd phicc c + di cho sdphicc a + bi khdc 0 Id tim sd phicc z sao cho c + di = (a + bi)z Sdphicc z dugc ggi la thuang trong phep chia

c + di cho a + bi vd ki hieu la

c + di

z =

• GV dua ra cac cau hdi sau:

H6 Tim z sao cho i = 2z

H7 Tim z sao cho i = (3i +3)z

H8 Ta da biet z.z = IzP = a^ + b^ Hay tim - va =

• GV neu cac budc tim thuong cua hai so phiic :

De tim thuong z = ta thuc hien cac budc sau:

a + bi

Budc 1 Dua vl dang (a + bi)z = c + di

Budc 2 Nhan ca hai ve vdi so phiic lien hgp ciia a +bi

Budc 3 z = — jUac +bd) + iad - bc)i]

a +b

• GV neu chii y :

Trong thuc hdnh, detinh thuang —, ta nhdn cd ?«" va mdu vdi sd

a + bi phicc lien hgp cua a + bi

H9 Cho sd phiic z = a + bi Tim z' ma z'.z = 1

• Cho sd phiic z = a + bi, sd phiic nghich dao ciia z la z'

a ^ + b ^ a ^ + b ^

• GV neu : Hai sd z va z' ggi la hai sd phirc nghich dao ciia nhau

HIO Chiing minh rdng : dl chia z cho z' ta nhan z vdi nghich dao ciia z'

• Thuc hien vi du 2 trong 5'

Ta cd z (2 + 3i) = 1 « z.l3 = 2-3i

2 3

z = 1

13 13

Ggi y tra Idi cau hdi 2

HS tu thuc hien bdng each nhan 3 + 2i vdi z

' Thuc hien "^C 2 trong 5

Hoat dgng ciia GV Hoat dgng cua HS

• GV cd the tong kit

MQT SO CflU HOI TRflC NGHIEM KHflCH QUflN

Hay dien dung sai vdo d trdng sau:

Cdu I Cho sd phiic z = 3 + 2i

(a) z = 3 - 2 i

(b) z + z = 2

(c) z.z = 13

3 2 (d) nghich dao cua z la i

(c) z.z = 13

3 2 (d) nghich dao ciia z la i

Hdy chgn khdng dinh dung trong cdc cdu sau:

Cdu 4 Trong cac so sau, sd nao la nghich dao cua sd phiic 2 + i

Cdu 5 Trong cac sd sau, sd nao la nghich dao ciia sd phiic 2 - i

HOATD0NG5

HCIGFNG DflN Bfll TflP SGK

Bai 1 Hudng ddn Six dung true tiep cdng thiic : -—^ = —^—— + - ^ — - 7 / c + di _ ac + bd ad-be

a + bi~ a^+b^ a^+b^

cau a Hudng ddn Nhan ca tii va mdu vdi 3 + 2i

§4 Phurcfng t r i n h bac h a i v d i h e s o thi;fc

( t i e t 8 )

I M U C TIEU

1 Kien thurc

HS ndm dugc :

Cach giai phuang trinh bac hai vdi he sd thuc

• Can bac hai ciia mdt s6' thuc am

2 KT nang

- Tim dugc cdng thiic nghiem cua phuang trinh bac hai

3 Thai do

- Tu giac, tich cue trong hgc tap

Biet phan biet rd cac khai niem co ban va van dung trong timg trudng hgp cu the

- Tu duy cac va'n dl ciia toan hgc mdt each Idgic va he thdng

II C H U A N BI CUA GV VA HS

1 Chuan bj cua GV

Chuan bi cac cau hdi ggi md

• Chudn bi pha'n mau, va mdt sd dd dung khac

IV TIEN TRINH DAY HOC

B BAI MOI

3 - i b) Giai phuong trinh (4-3i)z = i

HOAT DONG 1

1 Can bac hai ciia so thuc am

• Thuc hien -jpr / trong 5'

HS tu nhdc lai

Ggi y tra Idi cau hdi 2

HS tu tra Idi

HI Ne'u a = b thi a la can bac hai ciia b Diing hay sai?

H2 Neu a = b^ thi -a la can bac hai ciia b Diing hay sai?

• GV neu dinh nghia

(a) i ^ ; (a) - ^^3 ;

H3 can bac 2 ciia - 3 la

(a) ± i ^ • (b) ± i ^ ;

H5 Can bac 2 ciia - 5 la

(a) + / ^ ; (b) ± / ^ ; (a) i/s ; (a) - iyfs

GV cho HS tu dat va thuc hien cac phep tinh sau de hinh thanh kl nang:

a < 0

>

HOAT DONG 2

2 Phuong trinh bac hai vdi he sd thuc

H6 Trong cac phuang trinh sau, phuang trinh nao khdng cd nghiem thuc?

(a) J C ^ + 1 1 0 0 A ; - 1 = 0 ;

(a) j c ^ - 1 1 0 0 x - l = 0 ;

(b) x^+1100x + l = 0 ; (a) x^ + x + l = 0

GV neu van de : Cho phuang trinh bac hai ax + 6;c + c = 0 vdi a, 6, c e M , a ^ 0

A = 6 ^ - 4 a c

Ggi y tra Idi cau hdi 2

A = 0, phuang trinh cd mdt nghiem

b

t h u c x =

2a