Question

[Nguyễn Bảo Châm]

giúp mình với 

Answer 1

[Hoàng Lê Bảo Ngọc]

Ta có : \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{x+y+z}{2}\)(ĐKXĐ : \(x\ge0;y\ge1;z\ge2\))

\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)

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

Mà \(\left(\sqrt{x}-1\right)^2\ge0\); \(\left(\sqrt{y-1}-1\right)^2\ge0\) ; \(\left(\sqrt{z-2}-1\right)^2\ge0\)

Suy ra : (1) \(\Leftrightarrow\begin{cases}\left(\sqrt{x}-1\right)^2=0\\\left(\sqrt{y-1}-1\right)^2=0\\\left(\sqrt{z-2}-1\right)^2=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=2\\z=3\end{cases}\) (TMĐK)

Vậy \(\left(x_0;y_0;z_0\right)=\left(1;2;3\right)\)

\(S=x_0^2+y_0^2+z_0^2=1^2+2^2+3^2=14\)