Question

Tìm x,y,z,

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

Answer 1

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