Question

S=\(\left(\dfrac{x^3-3x}{x^2-9}-1\right):\left[\dfrac{9-x^2}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-3}{x+3}-\dfrac{x+2}{x-2}\right]\)

Answer 1

\(S=\left(\dfrac{x^3-3x}{x^2-9}-1\right):\left[\dfrac{9-x^2}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-3}{x+3}-\dfrac{x+2}{x-2}\right]\)

\(=\left[\dfrac{x\left(x^2-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left[\dfrac{\left(3-3\right)\left(3+x\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\right]\) Kiểu sai đề á >.<