问题标题：
先化简再求值（x+1分之x+x^2-1分之x+1)除以(x^2+x分之x^2+1)如题..
问题描述：
先化简再求值（x+1分之x+x^2-1分之x+1)除以(x^2+x分之x^2+1)
如题..
常富洋回答：
（x+1分之x+x^2-1分之x+1)除以(x^2+x分之x^2+1)
=（x/x+1+x+1/(x+1)(x-1))/(x^2+1/x^2+x)
=（x/x+1+1/x-1)*(x^2+x/x^2+1)
=（(x(x-1)+1)/(x+1)(x-1))*(x^2+x/(x^2+1))
=（(x^2-x+1)/(x+1)(x-1))*(x(x+1)/(x^2+1))
=x(x^2-x+1)/(x-1)(x^2+1)
=(x^3-x^2+x)/(x^3+x-x^2-1)
=(x^3-x^2+x-1+1)/(x^3-x^2+x-1)
=1+1/(x^3-x^2+x-1)
再将x的值代入,就可计算了.
齐梅回答：
但他的答案怎么是x-1分之x。。。。
常富洋回答：
不好意思，算错了。应是（x+1分之x+x^2-1分之x+1)除以(x^2+x分之x^2+1)=（x/x+1+x+1/(x+1)(x-1))/(x^2+1/x^2+x)=（x/x+1+1/x-1)*(x^2+x/x^2+1)=（(x(x-1)+x+1)/(x+1)(x-1))*(x^2+x/(x^2+1))=（(x^2-x+x+1)/(x+1)(x-1))*(x(x+1)/(x^2+1))=x(x^2+1)/(x-1)(x^2+1)=x/(x-1)
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