Question

Giúp mk vs ạ

Answer 1

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

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

\(=\dfrac{x+1}{x-3}\)

b: \(B=\left(\dfrac{4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-2}{x\left(x+2\right)}-\dfrac{x}{2\left(x+2\right)}\right)\)

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

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

\(=\dfrac{-2}{x-2}\)