Đặt $\begin{cases}u=x \\ v=x+y\\w=y+2z \end{cases}$ Hệ $\Leftrightarrow \begin{cases}uvw=\frac{1}{8} \\ uv+vw+wu=-\frac{3}{4} \\u+v+w=0\\u<v<w\end{cases}$
( Chú ý: $x^2+y^2+3xy+4xz+2yz=x(x+y)+(x+y)(y+2z)+$
$(y+2z)x$
$\Rightarrow u,v,w$ là nghiệm phương trình:
$ X^3-\frac{3}{4}X-\frac{1}{8}=0 \Leftrightarrow 4X^3-3X=\frac{1}{2}$
Đặt $
X=cos\alpha,\alpha\in[0;\pi]$
PT
$\Leftrightarrow 4cos^3\alpha-3cos\alpha=\frac{1}{2}$
$\Leftrightarrow cos3\alpha=\frac{1}{2}$
$\Leftrightarrow
\alpha\in ${$\frac{\pi}{9};\frac{5\pi}{9};\frac{7\pi}{9}$}
$\Leftrightarrow \left[ \begin{array}{l}X = \cos \frac{\pi}{9}\\X =\cos \frac{5\pi}{9} \\X=\cos \frac{7\pi}{9}\end{array} \right.$
Mà $u<v<w$
$\Leftrightarrow \begin{cases}u =cos \frac{7\pi}{9}\\v =\cos \frac{5\pi}{9}\\w=\cos \frac{\pi}{9} \end{cases}$
$\Leftrightarrow \begin{cases}x = \cos \frac{7\pi}{9}\\y = \cos \frac{5\pi}{9}-\cos \frac{7\pi}{9}\\z=\frac{1}{2}(\cos \frac{\pi}{9}-\cos \frac{5\pi}{9}+\cos \frac{7\pi}{9})\end{cases} .$