F.4maths續多項式

1.
1/(x-2)(x+2) - 1/(x+4)(x-2)

=[(x+4)-(x+2)] / [(x-2)(x+2)(x+4)] 先通分母

= 2/[(x-2)(x+2)(x+4) 分子 x-x 冇左 4+(-2) = 4-2 = 2

2.

先由右邊做起

1/x+4 + P/4x+5

= [(4x+5)+P(x+4)] / [(x+4)(4x+5)] 先通分母

= [(4+P)x+(5+4P)] / [(x+4)(4x+5)] 分子相加 再分有x的一堆，冇x的一堆

所以
Qx+1/(x+4)(4x+5) ≡ [(4+P)x+(5+4P)] / [(x+4)(4x+5)]
分母一樣，可以抵消
Qx+1+0 = (4+P)x+(5+4P) +0 係方便你理解，唔洗寫既

1 = 5+4P
P = (1-5)/4 = -1

Q = 4+P
= 4+(-1)
= 3

希望幫到你~~