- ĐKXĐ : \(2x-1\ne0\)
=> \(x\ne\frac{1}{2}\)
- Ta có : \(\frac{4x^3+4x^2+7x-5}{2x-1}\)
\(=\frac{4x^3-2x^2+6x^2-3x+10x-5}{2x-1}\)
\(=\frac{2x^2\left(2x-1\right)+3x\left(2x-1\right)+5\left(2x-1\right)}{2x-1}\)
\(=\frac{\left(2x-1\right)\left(2x^2+3x+5\right)}{2x-1}\)
\(=2x^2+3x+5\)