Đáp án:
Giải thích các bước giải:
$\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right)$
$=\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}$
$=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}$
$=\dfrac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}$
$=-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}$
`=-6/x`
