Đáp án:
\[D\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {2x - y} \right)^2} + {\left( {y - 2} \right)^2} + \sqrt {{{\left( {x + z} \right)}^2}} = 0\\
\Leftrightarrow {\left( {2x - y} \right)^2} + {\left( {y - 2} \right)^2} + \left| {x + z} \right| = 0\,\,\,\,\,\left( 1 \right)\\
{\left( {2x - y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\,\,\forall y\\
\left| {x + z} \right| \ge 0,\,\,\,\forall x,z\\
\Rightarrow {\left( {2x - y} \right)^2} + {\left( {y - 2} \right)^2} + \left| {x + z} \right| \ge 0,\,\,\,\forall x,y,z\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {2x - y} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0\\
\left| {x + z} \right| = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - y = 0\\
y - 2 = 0\\
x + z = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \frac{y}{2}\\
y = 2\\
z = - x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2\\
z = - 1
\end{array} \right.\\
\Rightarrow x + y + z = 1 + 2 + \left( { - 1} \right) = 2
\end{array}\)
