Question

giúp em vs ạ

Answer 1

\(ĐK:x\ne\pm3;x\ne0\\ M=\dfrac{\left(x+3\right)^2-\left(2x+5\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{1-x-3}{x\left(x+3\right)}\\ M=\dfrac{x^2+6x+9-2x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x+3\right)}{-\left(x+2\right)}\\ M=\dfrac{x^2+4x+4}{x-3}\cdot\dfrac{x}{-\left(x+2\right)}=\dfrac{x\left(x+2\right)^2}{3\left(3-x\right)}\\ M=\dfrac{x\left(x+2\right)^2}{\left(3-x\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3-x}\)