关于分式的题（x-1)[1-(x+1)/(x-1/x)]

原式=（x-1)[1-(x+1)*(x/x-1)]
=（x-1)[1-(x+1)x/（x-1)]
=（x-1)-(x+1)x 【把（x-1)乘进去中括号的每一项
=x-1-x²-x
=-x²-1

（x-1)[1-(x+1)/(x-1/x)]
=（x-1)[1-(x+1)/(x+1)(x-1)/x]
=（x-1)[1-(x+1)*x/(x+1)(x-1)]
=（x-1)[1-x/x-1]
=（x-1)*1-(x-1)*(x/x-1)
=(x-1)-x
=-1
这个够不够详细？

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

（x-1)[1-(x+1)/(x-1/x)]
=（x-1)[1-(x+1)/(x+1)(x-1)/x]
=（x-1)[1-(x+1)*x/(x+1)(x-1)]
=（x-1)[1-x/x-1]
=（x-1)*1-(x-1)*(x/x-1)
=(x-1)-x
=-1

（X-1）[1-(X+1)/(X-1/X)]= （X-1）[ 1- (X+1) (X/(X²-1)]= （X-1）[1-(X²+X)/(X²-1)]= （X-1）｛(X²-1-X²-X)/(X²-1)= （X-1）｛(-X-1)/[(X+1)(X-1)]｝= (-(X+1) ｛(X-1))/[(X+1)(X-1...