Question

先化简,再求值 {x-x/(x+1)}除以{1+1/(x^2-1)},其中x=(根号2)+1？

Answer 1

{x-x/(x+1)}/{1+1/(x^2-1)}
=x*[1-1/(x+1)]/[1+1/(x^2-1)]
=x*x(x-1)/x^2
=x-1=根号2,7,原式=[x^2/(x+1)]/[x^2/(x^2-1)]
=[x^2/(x+1)][(x^2-1)/x^2]
=(x^2-1/(x+1)
=x-1,2,原式=[x^2/(x+1)]/[x^2/(x^2-1)]
=[x^2/(x+1)][(x^2-1)/x^2]
=(x^2-1/(x+1)
=x-1
因为x=根号2+1
所以x-1=根号2,1,原式等于{x/(x+1)}/{x的平方/(x的平方-1)}=（x的平方-1）/(x+1)=x-1,,,,,,,,,,,,,,,,,,,,,,,,,,,,然后x=根号2+1，，，，最后的答案为根号2,1,