Question

x+1分之1 + x-1分之1 +(x-1)²分之1 怎么解啊 ？？跪求

Answer 1

1/（x+1)+1/(x-1)+1/(x-1)²
=[(x-1)²+(x-1)(x+1)+(x+1)]/[(x-1)²(x+1)]
=(x²-2x+1+x²-1+x+1)/[(x-1)²(x+1)]
=(2x²-x+1)/[(x-1)²(x+1)]
=[(2x+1)(x-1)]/[(x-1)²(x+1)]
=(2x+1)/[(x-1)(x+1)]
=(2x+1)/(x²-1)

追问

你的倒数第3个等号 （2x+1）（x-1）的出来的结果好像是 2x²-x-1啊 不是 2x²-x+1

追答

谢谢你的指正，那就只能算到倒数第三步了。即：
1/（x+1)+1/(x-1)+1/(x-1)²
=[(x-1)²+(x-1)(x+1)+(x+1)]/[(x-1)²(x+1)]
=(x²-2x+1+x²-1+x+1)/[(x-1)²(x+1)]
=(2x²-x+1)/[(x-1)²(x+1)]