求a，b的值。
更新1:
唔係2/x^2≡a/x+1 + b/x-1 而係2/x^2-1≡a/x+1 + b/x-1 求a，b的值。 sorry!!
數學恆等式。(!急!)
回答 (2)
2007-08-16 10:19 pm
✔ 最佳答案
Partial Fraction2/x^2-1≡a/x+1 + b/x-1
2/(x+1)(x-1)≡[a(x-1)+b(x+1)]/(x+1)(x-1)
So, 2=a(x-1)+b(x+1)
When x=-1, we get a=-1.
When x=1, we get b=1.
2007-08-14 5:43 am
2/x^2≡a/x+1 + b/x-1
2x^2(x+1)(x-1)/x^2≡ax^2(x+1)(x-1)/x+1 + bx^2(x+1)(x-1)/x-1
2x^2-2≡ax^2(x-1) + bx^2(x+1)
b-a = - 2
(a+b)x=2
a+b=2/x
2b=2/x - 2
b=1/x -1
a=1/x+1
2x^2(x+1)(x-1)/x^2≡ax^2(x+1)(x-1)/x+1 + bx^2(x+1)(x-1)/x-1
2x^2-2≡ax^2(x-1) + bx^2(x+1)
b-a = - 2
(a+b)x=2
a+b=2/x
2b=2/x - 2
b=1/x -1
a=1/x+1
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