Question

Tính

\(\dfrac{x^2}{2x^2+8x+8}.\dfrac{3x+6}{-x}\)

Answer 1

\(\dfrac{x^2}{2x^2+8x+8}\cdot\dfrac{3x+6}{-x}\left(dkxd:x\ne0;x\ne-2\right)\)

\(=\dfrac{x}{2\left(x^2+4x+4\right)}\cdot\dfrac{3x+6}{-1}\)

\(=\dfrac{x\cdot3\left(x+2\right)}{-2\left(x+2\right)^2}\)

\(=\dfrac{3x}{-2\left(x+2\right)}=\dfrac{-3x}{2x+4}\)