Đặt \(\sqrt[3]{x^2+5x-2}=y\)
\(pt\Leftrightarrow y^3-2y+4=0\)
\(\Leftrightarrow\left(y+2\right)\left(y^2-2y+2\right)=0\)
\(\Leftrightarrow y=-2\left(\text{Vì }y^2-2y+2>0\right)\)
\(\Leftrightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy ...