Q1. 1/(1-x^2)-1/(1+x)^2
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2007-03-15 4:42 am
回答 (2)
2007-03-15 5:19 am
✔ 最佳答案
Q1. 1/(1-x^2)-1/(1+x)^2＝ 1/(1-x²)-1/(1+x)²
＝ 1/[(1-x)(1+x)] - 1(1+x)²
＝ [ 1/(1+x) ] [ 1/(1-x) - 1/(1+x) ]
＝ [ 1/(1+x) ] { [(1+x) - (1-x)] / [(1-x)(1+x)]}
＝ [ 1/(1+x) ] { 2x / [(1-x)(1+x)]}
＝ 2x/ [(1+x)²(1-x)]
2007-03-15 5:34 am
Q1. 1/(1-x^2)-1/(1+x)^2＝ 1/(1-x²)-1/(1+x)²
＝ 1/[(1-x)(1+x)] - 1(1+x)²
＝ [ 1/(1+x) ] [ 1/(1-x) - 1/(1+x) ]
＝ [ 1/(1+x) ] { [(1+x) - (1-x)] / [(1-x)(1+x)]}
＝ [ 1/(1+x) ] { 2x / [(1-x)(1+x)]}
＝ 2x/ [(1+x)²(1-x)]
＝ 1/[(1-x)(1+x)] - 1(1+x)²
＝ [ 1/(1+x) ] [ 1/(1-x) - 1/(1+x) ]
＝ [ 1/(1+x) ] { [(1+x) - (1-x)] / [(1-x)(1+x)]}
＝ [ 1/(1+x) ] { 2x / [(1-x)(1+x)]}
＝ 2x/ [(1+x)²(1-x)]
參考: Oh yeah
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