ĐK: \(x\ne0;x\ne3\)
Khi đó:
\(\dfrac{1}{x}+\dfrac{2}{x-3}-\dfrac{6}{x^2-3x}\\ =\dfrac{1\left(x-3\right)}{x\left(x-3\right)}+\dfrac{2.x}{x\left(x-3\right)}-\dfrac{6}{x\left(x-3\right)}\\ =\dfrac{x-3+2x-6}{x\left(x-3\right)}\\ =\dfrac{3x-9}{x\left(x-3\right)}\\ =\dfrac{3\left(x-3\right)}{x\left(x-3\right)}\\ =\dfrac{3}{x}\)