Question

Cho : x²+y²+z²+3=2(x+y+z)

CMR:x=y=z=1

 

 

Answer 1

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-1=0 \\y-1=0\\z-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-1=0\\y-1=0\\z-1=0\end{cases}}\)

Vậy \(x=y=z=1\)