Question

Rút gọn biểu thức \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)

Answer 1

Lời giải:
ĐKXĐ: $x\neq 0;1$

\(A=\frac{x(x+1)}{(x-1)^2}:\left[\frac{(x+1)(x-1)}{x(x-1)}+\frac{1}{x-1}+\frac{2-x^2}{x(x-1)}\right]\)

\(=\frac{x(x+1)}{(x-1)^2}:\left[\frac{x^2-1}{x(x-1)}+\frac{x}{x(x-1)}+\frac{2-x^2}{x(x-1)}\right]\)

\(=\frac{x(x+1)}{(x-1)^2}:\frac{x^2-1+x+2-x^2}{x(x-1)}=\frac{x(x+1)}{(x-1)^2}:\frac{x+1}{x(x-1)}=\frac{x(x+1)}{(x-1)^2}.\frac{x(x-1)}{x+1}=\frac{x^2}{x-1}\)