Cau hdi 1
Tim so phirc nghich dao cua cac so phiic sau :
1 2 - i ; i; -3-12i. Cau hdi 2 Cau hdi 2
Thuc hien cac phep tinh sau : 12-i 12-i
a) Tinh
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'
Hoat dgng ciia GV Cau hdi 1
Dinh nghia can bac 2 ciia so duang a.
Cau hdi 2
Tim can bac hai ciia 7
Hoat dgng ciia HS Ggi y tra Idi cau hdi 1
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
•2
/ --\ ,ta noi i Id mot cdn bac hai cua -1. Cdng vdy, -i cUng Id mot cdn bdc hai cua -1. cdn bdc hai cua -1.
(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
>
J~a b < 0 VU yla + h
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.
Hoat dgng ciia GV Cau hdi 1
Tfnh A.
Cau hdi 2
A = 0, phuang trinh cd
nghiem nhu the' nao?
Hoat dgng cua H S Ggi y tra Idi cau hdi 1
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 =
C a u hdi 3
A > 0, phuang trinh cd nghiem nhu the' nao?
C a u hdi 4
A < 0, phuang trinh cd nghiem nhu the' nao?
Ggi y t r a Idi c a u hdi 3
- 6 + V A
x\a -'
2a
Ggi y t r a Idi c a u hdi 4
K,hi A < 0 phuong trinh khdng cd nghiem thuc.
• GV neu kit luan :
A < 0, ne'u xet trong tap hgp sdphicc, ta vdn co hai cdn bdc hai do cua
A la ± /^|AJ Khi do, phuang trinh co hai nghiem phicc dugc xdc dinh bdi cong thicc JC] 2 =
• Thuc hien vf du trong 5'.
-b ± iyl\A 2a Hoat dgng ciia 'GV Cau hdi 1 Tinh A. Cau hdi 2 Tim nghiem. Hoat dgng ciia HS Ggi y tra Idi cau hdi 1
A = 1 - 4 = - 3 .
Ggi y tra Idi cau hdi 2
-1±/V3
' Chii y trong SGK cd the neu nhimg day khdng phai la chii y quan trgng. Thuc hien mdt so vf du khac (GV tu ra d l )
Hoat dgng ciia GV Cau hdi 1
Giai phuang trinh : x2+3 = 0
Hoat dgng ciia HS Ggi y tra Idi cau hdi 1
Cau hdi 2
Giai phuang trinh :
x ^ + x + 3 = 0.
Ggi y tra Idi cau hdi 2
- l + iVlT
X = •
• GV cd the tong ket:
Chii y : Chiing ta cd thi sir dung A' : x^2 _-b'±i^' a a
• Thuc hien mdt sd vi du khac (GV tU ra
Hoat dgng ciia G V Cau hdi 1
Giai phuang trinh bdng each tinh A': x 2 + 3 = 0
Cau hdi 2
Giai phuang trinh bdng each tfnh A': x ^ - 2 x + 3 = 0.
d l )
Hoat dgng cua HS Ggi y tra Idi cau hdi 1
x= ±iV3
Ggi y tra Idi cau hdi 2
1
HOAT DONG 3
TOM TflT Bfll HPC
1. Can bac hai cua sd thuc a < 0 la + iJ \a\
2. Phuang trinh cd hai nghiem phiic dugc xac dinh bdi cdng thiic
-b±i4
Xl2 = 2a
3. Phuang trinh cd hai nghiem phiic dugc xac dinh bdi cong thiic
-b'±iyl\A'
HOAT DONG 4
MpT SO CflU H 6 I TR^C NGHIEM KHflCH QUflN
Hdy dien dung sai vdo 6 trdng sau Cdu 1. (a) Vs la ±iV5 (b) V ^ la ±iV5 (c) yp25 la ±5i (d) V ^ la iVs Trd Idi (a) s (b) D (c) D (d) s
Cdu 2. Phuang trinh x^ - 2x + 7 = 0
(a) Khdng cd nghiem thuc
(b) Cd hai nghiem phiic
(c) Khdng cd nghiem
(d) Ca ba kit luan tren sai
Trd Idi (a) D (b) D (c) S (d) S
Cdu 3. Cho so phiic x - 1 = 0
(a) Cd mdt nghiem (b) Cd mdt nghiem thuc D D D D D D D 0 D D
(c) Co 3 nghiem
(d) Ca ba kit luan tren sai
Trd Idi (a) S (b) D (c) D (d) S D D
Hdy chgn khdng dinh dung trong cdc cdu sau: Cdu 4. Phuang trinh x - 1 cd nghiem la
(a) 1 ;