10 当$x=$时，$6-\sqrt{2x+3}$有最大值，最大值为。

10. $-\dfrac{3}{2}\quad 6$

11 若$\sqrt{x+3}$的值与$\sqrt{2y-4}$的值互为相反数，求$(2x-3y+10)^2$的平方根.

11. 根据题意,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$\begin{cases} x=-3, \\ y=2, \end{cases}$ $\therefore (2x-3y+10)^2=4.$ $\because 4$ 的平方根为$±2,\therefore (2x-3y+10)^2$ 的平方根为$±2$

12 若实数 $a,b$ 满足 $|a - b + 1| = -\sqrt{a + 2b + 4}$，求 $3a + 3b$ 的立方根。

12. 根据题意,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$\sqrt[3]{3a+3b}=\sqrt[3]{-9}=-\sqrt[3]{9}$

13 若实数$x,y,z$满足$\frac{1}{2}|x-y| + z^2 + \frac{1}{4} - z + \sqrt{2y + z} = 0$，求$x(y+z)$的值。

13. 根据题意,得 $\dfrac{1}{2}\ |x-y|\ +(z-\dfrac{1}{2})^2+\sqrt{2y+z}=0.$ $\because \dfrac{1}{2}\ |x-y|≥0,(z-\dfrac{1}{2})^2≥0,\sqrt{2y+z}≥0,\therefore x-y=0,$ $z-\dfrac{1}{2}=0,2y+z=0.$ 联立,解得$\begin{cases} x=-\dfrac{1}{4}, \\ y=-\dfrac{1}{4}, \\ z=\dfrac{1}{2}. \end{cases}$ $\therefore x(y+z)= -\dfrac{1}{4}×(-\dfrac{1}{4}+\dfrac{1}{2})=-\dfrac{1}{16}$