10.若实数$x,y,z$满足条件$\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{4}(x+y+z+9)$，求$xyz$的值.

10.解:根据题意,得$x-4\sqrt{x}+y-4\sqrt{y-1}+z-4\sqrt{z-2}+9=0$,
$\therefore(x-4\sqrt{x}+4)+(y-1-4\sqrt{y-1}+4)+(z-2-4\sqrt{z-2}+4)=0$,
即$(\sqrt{x}-2)^2+(\sqrt{y-1}-2)^2+(\sqrt{z-2}-2)^2=0$,
$\therefore\sqrt{x}-2=0$且$\sqrt{y-1}-2=0$且$\sqrt{z-2}-2=0$,
$\therefore\sqrt{x}=2,\sqrt{y-1}=2,\sqrt{z-2}=2$,
$\therefore x=4,y-1=4,z-2=4$,
解得$x=4,y=5,z=6$.
$\therefore xyz=120$.

11.已知$\frac{\sqrt{3a - b} + |a^2 - 49|}{\sqrt{a + 7}} = 0$，求实数$a,b$的值，并求出$\sqrt{b}$的整数部分和小数部分。

11.解:根据题意,得$3a-b=0,a^2-49=0$且$a+7＞0$,
解得$a=7,b=21$.
$\because16＜21＜25,\therefore4＜\sqrt{21}＜5$,
$\therefore\sqrt{21}$的整数部分是4,小数部分是$\sqrt{21}-4$.

12. 对一切实数$k$，有$\sqrt{8-x}+\sqrt{x-3} ≥ k$总成立，求$k$的最大值.

12.解:要使根式有意义,则$3≤ x≤8$.
设$y=\sqrt{8-x}+\sqrt{x-3}$,
则$y^2=8-x+x-3+2\sqrt{(8-x)(x-3)}=5+2\sqrt{-(x-\frac{11}{2})^2+\frac{25}{4}}$,
$\therefore$当$x=\frac{11}{2}$时,$y^2$有最大值,最大值为$5+2×\frac{5}{2}=10$,
$\therefore y$的最大值为$\sqrt{10}$,即$\sqrt{10}≥ k$,
$\therefore k$的最大值为$\sqrt{10}$.