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

       \(\left(x+2y\right)^2=0\Leftrightarrow x+2y=0\Leftrightarrow x=2y\left(1\right)\)

       \(\left(y-1\right)^2=0\Leftrightarrow y-1=0\Leftrightarrow y=1\left(2\right)\)

          \(\left(x-z\right)^2=0\Leftrightarrow x-z=0\Leftrightarrow x=z\left(3\right)\)

 \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow2y=x=y=2\left(4\right)\)

                      \(\left(4\right)\Leftrightarrow A=2+2+3\times2=10\)

\(\left(x+2y\right)^2\ge0;\left(y-1\right)^2\ge0;\left(x-z\right)^2\ge0\)

\(\Rightarrow\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2\ge0\)

theo đề:\(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)

\(\Rightarrow\left(x+2y\right)^2=\left(y-1\right)^2=\left(x-z\right)^2=0\)

+)y-1=0=>y=1

ta có:x+2y=0=>x+2=0=>x=-2

Mà x-z=0=>x=z=>z=-3

Vậy x+2y+3z=(-2)+2+3.(-3)=3.(-3)=-27

Ta có : \(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}y-1=0\\x+2y=0\\x-z=0\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x+2.1=0\\x-z=0\end{cases}\Rightarrow}\hept{\begin{cases}y=1\\x=-3\\\left(-3\right)-z=0\end{cases}\Rightarrow}\hept{\begin{cases}y=1\\x=-3\\z=-3\end{cases}}}\)

Ta có : \(\hept{\begin{cases}y=1\\x=-3\\z=-3\end{cases}}\)

Bạn thế vào : \(x+2y+3z\)là ra thôi 

giải giúp mình nha 

\(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+2y=0\\y-1=0\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y=1\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\\z=-2\end{cases}}\)

Do đó: \(x+2y+3\text{z}=-2+2-2.3=-6\)

Vậy: \(M=-6\)

Vì (x+2y)² ; (y-1)² ; (x-z)²  lớn hơn hoặc bằng 0

=>x+2y = 0 ; y-1=0 ;x-z=0

=>x=-2y ; y=1 ;z=x

=>x=-2 ; y=1 ; z=-2