Question

X^2+2y^2+z^2-2(xy+2y+2z+8)=0

 

Answer 1

\(x^2+2y^2+z^2-2\left(xy+2y+2z+8\right)=0\)

\(pt\Leftrightarrow x^2+2y^2+z^2-2xy+4y+4z+16=0\)

\(\Leftrightarrow x^2-2xy+y^2+y^2+4y+4+z^2+4z+4+8=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8=0\)

Dễ thấy: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y+2\right)^2\ge0\\\left(z+2\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8>8\)

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