Question

Tìm x, y, z biết:

\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)

Answer 1

ĐKXĐ: \(x\ge-1;y\ge3;z\ge1\)

\(\Leftrightarrow x+y+z-2\sqrt{x+1}-2\sqrt{y-3}-2\sqrt{z-1}=0\)

\(\Leftrightarrow\left(x+1-2\sqrt{x+1}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+\left(z-1-2\sqrt{z-1}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}-1=0\\\sqrt{y-3}-1=0\\\sqrt{z-1}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)