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$-\frac{3}{2}$
6
解：由题意得$\sqrt{x+3} + \sqrt{2y - 4} = 0。$
$\because \sqrt{x+3} ≥ 0，$$\sqrt{2y - 4} ≥ 0，$
$\therefore \sqrt{x+3} = 0，$$\sqrt{2y - 4} = 0，$
即$\begin{cases} x + 3 = 0 \\ 2y - 4 = 0 \end{cases}，$
解得$\begin{cases} x = -3 \\ y = 2 \end{cases}。$
$\therefore (2x - 3y + 10)^2 = (-6 - 6 + 10)^2 = 4。$
$\because 4$的平方根为$\pm 2，$
$\therefore (2x - 3y + 10)^2$的平方根为$\pm 2。$
解：由题意得$|a - b + 1| + \sqrt{a + 2b + 4} = 0。$
$\because |a - b + 1| ≥ 0，$$\sqrt{a + 2b + 4} ≥ 0，$
$\therefore |a - b + 1| = 0，$$\sqrt{a + 2b + 4} = 0，$
即$\begin{cases} a - b + 1 = 0 \\ a + 2b + 4 = 0 \end{cases}，$
解得$\begin{cases} a = -2 \\ b = -1 \end{cases}。$
$\therefore 3a + 3b$的立方根为$\sqrt[3]{3a + 3b} = \sqrt[3]{-9} = -\sqrt[3]{9}。$
解：由题意得$\frac{1}{2}|x - y| + (z - \frac{1}{2})^2 + \sqrt{2y + z} = 0。$
$\because \frac{1}{2}|x - y| ≥ 0，$$(z - \frac{1}{2})^2 ≥ 0，$$\sqrt{2y + z} ≥ 0，$
$\therefore x - y = 0，$$z - \frac{1}{2} = 0，$$2y + z = 0。$
联立解得$\begin{cases} x = -\frac{1}{4} \\ y = -\frac{1}{4} \\ z = \frac{1}{2} \end{cases}，$
$\therefore x(y + z) = -\frac{1}{4} × (-\frac{1}{4} + \frac{1}{2}) = -\frac{1}{16}。$
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