Question

Tính

\(\dfrac{x^2-4x+4}{x^2+2x+1}.\dfrac{x+1}{x^2-2x}.\dfrac{6x}{2x+4}\)

Answer 1

\(\dfrac{x^2-4x+4}{x^2+2x+1}.\dfrac{x+1}{x^2-2x}.\dfrac{6x}{2x+4}=\dfrac{\left(x-2\right)^2}{\left(x+1\right)^2}.\dfrac{x+1}{x\left(x-2\right)}.\dfrac{6x}{2\left(x+2\right)}\)

\(=\dfrac{3\left(x-2\right)}{\left(x+1\right)\left(x+2\right)}\)