Question

Cần gấp ạ

Answer 1

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