Đáp án: $N = 2$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x \ge 2\\
y \le 1\\
z \ge - 3
\end{array} \right.\\
4x - y + z + 10 = 4\sqrt {x - 2} + 6\sqrt {1 - y} + 4\sqrt {z + 3} \\
\Leftrightarrow 4x - 4\sqrt {x - 2} - y - 6\sqrt {1 - y} \\
+ z - 4\sqrt {z + 3} + 10 = 0\\
\Leftrightarrow 4x - 8 - 2\sqrt {4x - 8} .1 + 1\\
+ 1 - y - 6\sqrt {1 - y} + 9\\
+ z + 3 - 4\sqrt {z + 3} + 4 = 0\\
\Leftrightarrow {\left( {\sqrt {4x - 8} - 1} \right)^2} + {\left( {\sqrt {1 - y} - 3} \right)^2} + {\left( {\sqrt {z + 3} - 2} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {\sqrt {4x - 8} - 1} \right)^2} \ge 0\\
{\left( {\sqrt {1 - y} - 3} \right)^2} \ge 0\\
{\left( {\sqrt {z + 3} - 2} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sqrt {4x - 8} - 1} \right)^2} = 0\\
{\left( {\sqrt {1 - y} - 3} \right)^2} = 0\\
{\left( {\sqrt {z + 3} - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {4x - 8} = 1\\
\sqrt {1 - y} = 3\\
\sqrt {z + 3} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4x - 8 = 1\\
1 - y = 9\\
z + 3 = 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{9}{4}\\
y = - 8\\
z = 1
\end{array} \right.\left( {tmdk} \right)\\
\Leftrightarrow N = 4x + y + z\\
= 4.\dfrac{9}{4} + \left( { - 8} \right) + 1 = 2\\
Vậy\,N = 2
\end{array}$