Question

thực hiện phép tính:
\(\dfrac{x^2-5x+6}{x^2+7x+12}.\dfrac{x^2+3x}{x^2-4x+4}\)
 

Answer 1

ĐKXĐ: \(x\notin\left\{2;-3;-4\right\}\)

\(\dfrac{x^2-5x+6}{x^2+7x+12}\cdot\dfrac{x^2+3x}{x^2-4x+4}\)

\(=\dfrac{x^2-2x-3x+6}{x^2+3x+4x+12}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

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

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

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