ĐKXĐ của x là: \(x\ne-2;x\ne3\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x-1}{x-3}\Leftrightarrow\dfrac{x}{x+2}-\dfrac{x-1}{x-3}=0\Leftrightarrow\dfrac{x\left(x-3\right)-\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=0\Leftrightarrow x^2-3x-x^2-x+2=0\Leftrightarrow2-4x=0\Leftrightarrow x=\dfrac{1}{2}\)Vậy \(x=\dfrac{1}{2}\)