Question

[Nguyễn Thị Thùy Dung]

giải hpt sau:

\(\sqrt{x-2}+\sqrt{y+2012}+\sqrt{z-2013}=\dfrac{1}{2}\left(x+y+z\right)\)

Answer 1

[tran nguyen bao quan]

\(\sqrt{x-2}+\sqrt{y+2012}+\sqrt{z-2013}=\dfrac{1}{2}\left(x+y+z\right)\Leftrightarrow2\sqrt{x-2}+2\sqrt{y+2012}+2\sqrt{z-2013}=x+y+z\Leftrightarrow x-2\sqrt{x-2}+y-2\sqrt{y+2012}+z-2\sqrt{z-2013}=0\Leftrightarrow x-2-2\sqrt{x-2}+1+y+2012-2\sqrt{y+2012}+1+z-2013-2\sqrt{z-2013}+1=0\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2=0\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y+2012}-1=0\\\sqrt{z-2013}-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x-2=1\\y+2012=1\\z-2013=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=-2011\\z=2014\end{matrix}\right.\)

Vậy x=3;y=-2011;z=2014