1/x+1-x+3/x^2-1(x^2-2x+1/x^2+4x+3
答案:2 悬赏:20 手机版
解决时间 2021-11-17 12:11
- 提问者网友：战皆罪
- 2021-11-16 20:00
1/x+1-x+3/x^2-1(x^2-2x+1/x^2+4x+3
最佳答案
- 五星知识达人网友：雪起风沙痕
- 2021-11-16 20:09
原式=[1/(x+1)]-[(x+3)/(x²-1)]×[(x²-2x+1)/(x²+4x+3)]
=[1/(x+1)]-{ (x+3)/[(x+1)(x-1)] }×{ (x-1)²/[(x+1)(x+3)] }
=[1/(x+1)]-[(x-1)/(x+1)²]
=[(x+1)/(x+1)²]-[(x-1)/(x+1)²]
=[(x+1)-(x-1)]/(x+1)²
=(x+1-x+1)/(x+1)²
=2/(x+1)²
=[1/(x+1)]-{ (x+3)/[(x+1)(x-1)] }×{ (x-1)²/[(x+1)(x+3)] }
=[1/(x+1)]-[(x-1)/(x+1)²]
=[(x+1)/(x+1)²]-[(x-1)/(x+1)²]
=[(x+1)-(x-1)]/(x+1)²
=(x+1-x+1)/(x+1)²
=2/(x+1)²
全部回答
- 1楼网友：举杯邀酒敬孤独
- 2021-11-16 21:39
解：原式 =1/(x+1)-[(x+3)/(x²-1)]*(x²+2x+1)/(x²+4x+3)
=1/(x+1)-(x+3)/[(x-1)(x+1)]*(x+1)²/[(x+1)(x+3)]
=1/(x+1)-1/(x-1)
=2/(1-x²)
=1/(x+1)-(x+3)/[(x-1)(x+1)]*(x+1)²/[(x+1)(x+3)]
=1/(x+1)-1/(x-1)
=2/(1-x²)
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