Maths唔識抵死題!

( x^2 - 1 ) / ( x^3 - 2x^2 )
= (x+1)(x-1) / x^2 (x-2)
Let this be (Ax+B)/x^2 + C/(x-2)
(x+1)(x-1) / x^2 (x-2) = (Ax+B)/x^2 + C/(x-2)
(x+1)(x-1) = (Ax+B)(x-2) + Cx^2------(1)
When x = 0,
(1)(-1) = B(-2)
B = 1/2
When x = 2,
(3)(1) = 4C
C = 3/4
Sub B and C into (1),
(x+1)(x-1) = (Ax + 1/2)(x-2) + 3x^2/4
x^2 -1 = Ax^2 - 2Ax + x/2 - 1 + 3x^2/4
x^2 -1 = (A+3/4)x^2 - (2A - 1/2)x - 1
By comparing coefficient,
A+3/4 = 1------(2)
2A - 1/2 = 0--------(3),
From (2), A = 1/4

Therefore,
( x^2 - 1 ) / ( x^3 - 2x^2 )
= (x/4 + 1/2)/x^2 + 3/[4(x-2)]
= (1/4)(x+2)/x^2 + 3/[4(x-2)]

2010-10-01 20:17:34 補充：
sorry, I wrongly assume this as the answer.
as you have said the answer should be
(1/4)(x+2)/x^2 + 3/[4(x-2)]
= x/4x^2 + 1/2x^2 + 3/[4(x-2)]
= 1/4x + 1/2x^2 + 3/[4(x-2)]

the ans is wrong, it should be 1/2x^2 + 1/4x + 3/4(x-2)