e) $\sqrt{x^{2} + 2x} = -2x^{2} - 4x + 3$
$\Leftrightarrow x^{2} + 2x = 4x^{4} + 16x^{2} + 9 + 16x^{3} - 12x^{2} - 24x$
$\Leftrightarrow x^{2} + 2x = 4x^{4} + 4x^{2} + 9 + 16x^{3} - 24x$
$\Leftrightarrow x^{2} + 2x - 4x^{4} - 4x^{2} - 9 - 16x^{3} + 24x = 0$
$\Leftrightarrow -3x^{2} + 26x - 4x^{4} - 4x^{2} - 9 - 16x^{3} = 0$
$\Leftrightarrow 4x^{2} - 16x^{2} + 9x^{2} + 8x + 18x - 4x^{4} - 9 - 8x^{3} - 8x^{3} = 0$
$\Leftrightarrow -4x^{2}\left ( x^{2} + 2x - 1 \right ) - 8\left ( x^{2} + 2x - 1 \right ) + 9\left ( x^{2} + 2x - 1 \right ) = 0$
$\Leftrightarrow -\left ( x^{2} + 2x - 1 \right )\left ( 4x^{2} + 8x - 9 \right ) = 0$
$\Leftrightarrow \left ( x^{2} + 2x - 1 \right )\left ( 4x^{2} + 8x - 9 \right ) = 0$
$\Leftrightarrow \left[ \begin{array}{l}x^{2} + 2x - 1 = 0\\4x^{2} + 8x - 9 = 0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = -1 + \sqrt{2}\\x = -1 - \sqrt{2}\\x = \dfrac{-2 + \sqrt{13}}{2}\\x = \dfrac{-2 - \sqrt{13}}{2}\end{array} \right.$
Thử lại ta có: $\left[ \begin{array}{l}x = -1 + \sqrt{2}\\x = -1 - \sqrt{2} \end{array} \right.$ thỏa mãn phương trình
