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

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

\(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\\left(y-2\right)^2\ge0\\\left|x+y+z\right|\ge0\end{cases}\Rightarrow}\left(2x-y\right)^2+\left(y-2\right)^2+\left|x+y+z\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y\right)^2=0\\\left(y-2\right)^2=0\\x+y+z=0\end{cases}\hept{\begin{cases}2x=y\\y=2\\x+y+z=0\end{cases}\hept{\begin{cases}x=1\\y=2\\1+2+z=0\end{cases}\hept{\begin{cases}x=1\\y=2\\z=-3\end{cases}}}}}\)

vậy pt có nghiệm lần lượt (x,y,z) là (1,2,-3)

\(M=\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0\right)\)

Khi \(M=\sqrt{x}-2\)

\(\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}=x+\sqrt{x}-2\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}=x-\sqrt{x}-2\)

\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)

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

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=\left(\pm\sqrt{3}\right)^2\)

\(\Leftrightarrow\sqrt{x}-1=\pm\sqrt{3}\)

\(\Leftrightarrow\sqrt{x}=\pm\sqrt{3}+1\)

\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(-\sqrt{3}+1\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\1-2\sqrt{3}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\\x=4-2\sqrt{3}\end{cases}}\)

Vậy \(x\in\left\{4\pm2\sqrt{3}\right\}\)khi \(M=\sqrt{x}-2\)