Question

 làm tính nhân 2 phân thức

 

Answer 1

a)

\(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+2\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{\left(x-2\right).\left(x^2+4x\right)}{5x+10}\\ =\dfrac{x^3+4x^2-2x^2-8x}{5x+10}\\ =\dfrac{x^3+2x^2-8x}{5x+10}\)

b)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2x+10}.-\dfrac{3}{x-6}\\ =\dfrac{-3\left(x+6\right)}{2x+10}\\ =\dfrac{-3x-18}{2x+10}\)