\(\hept{\begin{cases}\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\left(1\right)\\x^2+z^2-4\left(y+z\right)+8=0\left(2\right)\end{cases}}\)
Ta có:(1) \(\Leftrightarrow\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)\left(\sqrt{y^2+2012}-y\right)\)\(=2012\left(\sqrt{y^2+2012}-y\right)\)(Do \(\sqrt{y^2+2012}-y\ne0\forall y\))
\(\Leftrightarrow2012\left(x+\sqrt{x^2+2012}\right)=2012\left(\sqrt{y^2+2012}-y\right)\)
\(\Leftrightarrow x+\sqrt{x^2+2012}=\sqrt{y^2+2012}-y\)\(\Leftrightarrow x+y=\sqrt{y^2+2012}-\sqrt{x^2+2012}\)
\(\Leftrightarrow x+y=\)\(\frac{\left(\sqrt{y^2+2012}+\sqrt{x^2+2012}\right)\left(\sqrt{y^2+2012}-\sqrt{x^2+2012}\right)}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)
\(\Leftrightarrow x+y=\frac{y^2-x^2}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)\(\Leftrightarrow\left(x+y\right)\frac{\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}=0\)
Do \(\hept{\begin{cases}\sqrt{y^2+2012}>\sqrt{y^2}=\left|y\right|\ge y\forall y\\\sqrt{x^2+2012}>\sqrt{x^2}=\left|x\right|\ge-x\forall x\end{cases}}\)\(\Rightarrow\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x>0\forall x,y\Rightarrow x+y=0\)
\(\Rightarrow y=-x\)
Thay y = -x vào (2), ta được: \(x^2+z^2+4x-4z+8=0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(z-2\right)^2=0\Leftrightarrow\hept{\begin{cases}x=-2\\z=2\end{cases}}\Rightarrow y=-x=2\)
Vậy hệ có nghiệm \(\left(x;y;z\right)=\left(-2;2;2\right)\)
