ĐK:...........PT $\Leftrightarrow \sqrt{x+1}+2=\frac{x^2-x-6}{\sqrt[3]{2x+1}-3}$
$\Leftrightarrow \frac{(x+2)(\sqrt{x+1}-2)}{\sqrt[2]{2x+1}-3}=1$
$\Leftrightarrow (2x+1)+\sqrt[3]{2x+1}=(x+1)\sqrt{x+1}+\sqrt{x+1}$
Xét hàm $f(t)=t^3+t$ đồng biến trên R
$\rightarrow $ PT $\Leftrightarrow \sqrt{x+1}=\sqrt[3]{2x+1}$
$\Rightarrow \left\{ \begin{array}{l} x\geq \frac{-1}{2}\\ (2x+1)^2=(x+1)^3 \end{array} \right.$
$\rightarrow ..............$