\(\frac{x^2-3x-x+3}{x-3}=\frac{x\left(x-3\right)-\left(x-3\right)}{x-3}=\frac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)( ĐK: \(x\ne3\))
\(\frac{2x^3-5x^2-4x+3}{2x-1}=\frac{\left(2x^3-x^2\right)-\left(4x^2-2x\right)-\left(6x-3\right)}{2x-1}=\frac{x^2\left(2x-1\right)-2x\left(2x-1\right)-3\left(2x-1\right)}{2x-1}=\frac{\left(2x-1\right)\left(x^2-2x-3\right)}{2x-1}=x^2-2x-3\)( ĐK: \(x\ne\frac{1}{2}\))
Tham khảo nhé~