Đặt \(x+1=z\Rightarrow x=z-1\) ta được:
\(\left\{{}\begin{matrix}4\left(z-1\right)y+z-1+4\sqrt{\left(3-z\right)\left(y+2\right)}=14\\z^2+y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4yz-4y+z+2\sqrt{\left(3-z\right)\left(4y+8\right)}=15\\2z^2+2y^2=4\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow2z^2+2y^2-4yz+4y-z-2\sqrt{\left(3-z\right)\left(4y+8\right)}=-11\)
\(\Leftrightarrow2\left(y-z\right)^2+\left(4y+8-2\sqrt{\left(3-z\right)\left(4y+8\right)}+3-z\right)=0\)
\(\Leftrightarrow2\left(y-z\right)^2+\left(\sqrt{4y+8}-\sqrt{3-z}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}y-z=0\\\sqrt{4y+8}=\sqrt{3-z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=z\\4y+8=3-z\end{matrix}\right.\)
\(\Rightarrow4y+8=3-y\)
\(\Rightarrow y=-1\Rightarrow z=-1\)
\(\Rightarrow x=z-1=-2\)
