中二級數學2條(須有詳細步驟)

1. L.H.S. = Ax + Bx(x+2) + C(x+2)^2
= Ax + Bx^2 + 2Bx + C(x^2 + 4x + 4)
= Bx^2 + Ax + 2Bx + Cx^2 + 4Cx + 4C
= (B+C)x^2 + (A+2B+4C)x + 4C
Hence, (B+C)x^2 + (A+2B+4C)x + 4C ≡ 2x^2 + 6x - 4
Comparing the coefficients,
We have,
B + C = 2 ---(1)
A + 2B + 4C = 6 ---(2)
4C = -4 --- (3)
From (3), C = -1
Put C = -1 into (1),
B + (-1) = 2
B = 3
Put B = 3 and C = -1 into (2),
A + 2(3) + 4(-1) = 6
A = 5

2. Comparing the coefficients,
4 = P - 1
P = 5
Q = -5

1)If Ax+Bx(x+2)+C(x+2)^2≡2x^2+6x－4，is an identity，then what are the values
of B and C ?
Sol
Ax+Bx(x+2)+C(x+2)^2≡2x^2+6x－4
When x=－2
－2A=8－12－4
A=4
4x+Bx(x+2)+C(x+2)^2≡2x^2+6x－4
Bx(x+2)+C(x+2)^2≡2x^2+2x－4
Bx+C(x+2)≡2x－2
When x=－2
－2B=－4－2
B=3
3x+C(x+2)=2x－2
C(x+2)=-x－2
C=－1

2)If 4x－5≡(P－1)x+Q，where P and Q are constants，find the values of P and Q.
Sol
4x－5=(P－1)x+Q
When x=0
－5=Q
4x－5=(P－1)x－5
P－1=4
P=5

1.A=2,B=6

2.P=4,Q=1