Question

Thu gọn

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

Answer 1

Ta có: \(\left(\dfrac{3}{x-1}+\dfrac{x+3}{x^2-1}\right):\left(\dfrac{x+2}{x^2+x-2}-\dfrac{x}{x+2}\right)\)

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

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

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

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