先化简，再求值
1个回答
展开全部
解：
x/(x-1)-x/(x²-1)+(x²-x)/(x²-2x+1)
=x(x+1)/[(x-1)(x+1)]-x/(x²-1)+x(x-1)/(x-1)²
=x(x+1)(x-1)/[(x-1)²(x+1)]-x(x-1)/[(x-1)²(x+1)]+x(x-1)(x+1)/[(x+1)(x-1)²]
=[x(x+1)(x-1)-x(x-1)+x(x-1)(x+1)]
x/(x-1)-x/(x²-1)+(x²-x)/(x²-2x+1)
=x(x+1)/[(x-1)(x+1)]-x/(x²-1)+x(x-1)/(x-1)²
=x(x+1)(x-1)/[(x-1)²(x+1)]-x(x-1)/[(x-1)²(x+1)]+x(x-1)(x+1)/[(x+1)(x-1)²]
=[x(x+1)(x-1)-x(x-1)+x(x-1)(x+1)]
追答
解：
x/(x-1)-x/(x²-1)+(x²-x)/(x²-2x+1)
=x(x+1)/[(x-1)(x+1)]-x/(x²-1)+x(x-1)/(x-1)²
=x(x+1)(x-1)/[(x-1)²(x+1)]-x(x-1)/[(x-1)²(x+1)]+x(x-1)(x+1)/[(x+1)(x-1)²]
=[x(x+1)(x-1)-x(x-1)+x(x-1)(x+1)]/[(x-1)²(x+1)]
=(x³-x-x²+x+x³-x)/[(x-1)²(x+1)]
=(2x³-x²-x)/[(x-1)²(x+1)]
=x(2x+1)(x-1)/[(x-1)²(x+1)]
=x(2x+1)/[(x-1)(x+1)]
在-1≤x≤3中取x=0代入上式得=0
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