Question

Tĩm x,y,z

x^2+5y^2+z^2+2yz=12y-2x-10

Answer 1

\(x^2+5y^2+z^2+2yz-12y+2x+10=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2yz+z^2\right)+\left(4y^2-12y+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y+z\right)^2+\left(2y-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+z=0\\2y-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\z=-\frac{3}{2}\\y=\frac{3}{2}\end{matrix}\right.\)