$ (x^2+4x+8)^2 +3x . (x^2 +4x +8) + 2x^2 = 0$
$\to (x^2 +4x +8)^2 + 2. \dfrac{3}{2}x . (x^2 +4x +8) + \dfrac{9}{4}x^2 - \dfrac{9}{4}x^2 + 2x^2 = 0$
$ \to (x^2 +4x +8 + \dfrac{3}{2}x)^2 - \dfrac{1}{4}x^2 = 0$
$ \to (x^2 + \dfrac{11}{2}x + 8)^2 = (\dfrac{1}{2}x)^2$
TH1
$x^2 + \dfrac{11}{2}x + 8 = \dfrac{1}{2}x$
$\to x^2 -5x +8 = 0$
$\to x^2 - 2. \dfrac{5}{2} x + \dfrac{25}{4} + \dfrac{7}{4} = 0$
$ \to (x - \dfrac{5}{2} )^2 + \dfrac{7}{4} = 0$
Ta có $ (x - \dfrac{5}{2} )^2 \ge 0$ nên $ (x - \dfrac{5}{2} )^2 + \dfrac{7}{4} >0$
$ \to$ Không có $x$ thỏa mãn
TH2
$x^2 + \dfrac{11}{2}x + 8 = -\dfrac{1}{2}x$
$\to x^2 +6x +8 = 0 \to (x+2)(x+4) = 0 \to$ \(\left[ \begin{array}{l}x+2=0\\x+4=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=-2\\x=-4\end{array} \right.\)
Vậy $ x \in \{ -4;-2\}$
