\(2x^2+2y^2+z^2+2xy+2yz+2xz+32x+34y+545=0\)
\(\Leftrightarrow\left(x^2+2.x.16^2+16^2\right)+\left(y^2+2.y.17+17^2\right)+\left(x^2+y^2+z^2+2xy+2yz+2zx\right)=0\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x+16\right)^2\ge0\forall z\\\left(y+17\right)^2\ge0\forall y\\\left(x+y+z\right)^2\ge0\forall x;y;z\end{matrix}\right.\)\(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2\ge0\forall x;y;z\)
Mà \(\Leftrightarrow\left(x+16\right)^2+\left(y+17\right)^2+\left(x+y+z\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+16\right)^2=0\\\left(y+17\right)^2=0\\\left(x+y+z\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+16=0\\y+17=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\x+y+z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=-17\\z-33=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-16\\y=17\\z=33\end{matrix}\right.\)
