Giúp em vs ạ $\frac{3}{2x +3}$ – $\frac{x – 6}{2x^{2}}$ + 6x

Điều kiện xác định $\left\{ \begin{array}{l} 2x + 3 \ne 0\\ 2{x^2} + 6x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne  – \dfrac{3}{2}\\ x \ne 0\\ x \ne  – 3 \end{array} \right.$

$\begin{array}{l} \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2{x^2} + 6x}}\\  = \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2x\left( {x + 3} \right)}}\\  = \dfrac{{3.2x\left( {x + 3} \right) – \left( {x – 6} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right).2x\left( {x + 3} \right)}}\\  = \dfrac{{6{x^2} + 18x – \left( {2{x^2} – 9x – 18} \right)}}{{\left( {2x + 3} \right).2x}}\\  = \dfrac{{4{x^2} + 27x + 18}}{{\left( {2x + 3} \right)2x.\left( {x + 3} \right)}} \end{array}$