Question

 giải phương trình: x^2+2z+y^2-2x+z^2-2y+3=0

 

 

Answer 1

\(x^2+2z+y^2-2x+z^2-2y+3=0\)

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

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

Do \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-1\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y;z\)

\(\Rightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x-1=0\\y-1=0\\z-1=0\end{matrix}\right.\) \(\Rightarrow x=y=z=1\)