Question

Nhân phân thức:
a)\(\dfrac{3-3x}{x^2-9}\).\(\dfrac{x-3}{x-1}\)
b)\(\dfrac{6x+4}{x^2-4}\).\(\dfrac{x^2-2x}{3x+2}\)

Answer 1

\(\dfrac{3-3x}{x^2-9}\cdot\dfrac{x-3}{x-1}\\ =\dfrac{3\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =\dfrac{-3\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =-\dfrac{3}{x+3}\\ \dfrac{6x+4}{x^2-4}\cdot\dfrac{x^2-2x}{3x+2}\\ =\dfrac{2\left(3x+2\right)x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}\\ =\dfrac{2x}{x+2}\)