(x^2-2x)(x^-1)^(-1)/[(x-1)—(2x-1·)(x+1)^(-2)],中x=二分之一怎么算呢
问题描述:
(x^2-2x)(x^-1)^(-1)/[(x-1)—(2x-1·)(x+1)^(-2)],中x=二分之一怎么算呢
答
(x^2-2x)/(x^2-1)÷{x-1-(2x-1)/(x+1)}
=x(x-2)/(x+1)(x-1)÷[(x^2-1-2x+1)/(x+1)]
=x(x-2)/(x+1)(x-1)÷[x(x-2)/(x+1)]
=x(x-2)/(x+1)(x-1)*(x+1)/x(x-2)
=1/(x-1)
=-2.