8. $m$、$n$均为整数，且满足$n = \sqrt{\frac{m}{2}-1}-\sqrt{3 - m}$，则$m + n$的值为 （ ）
A. 1
B. 2
C. 3
D. 4

A 解析：$\because n = \sqrt{\frac{m}{2}-1}-\sqrt{3 - m}$，$\therefore \begin{cases}\frac{m}{2}-1\geq0 \\ 3 - m\geq0\end{cases}$，解得$2\leq m\leq3$.
$\because m$为整数，$\therefore m = 2$或$m = 3$. 又$\because n$为整数，$\therefore m = 2$，此时$n = -1$，
$\therefore m + n = 1$. 故选 A.

9. 仿照$3x^{2}-9 = 3(x+\sqrt{3})(x - \sqrt{3})$，将下列式子在实数范围内分解因式：
(1)$4x^{4}-1=$______________；
(2)$x^{4}-4x^{2}+4=$______________.

(1)$(2x^2 + 1)(\sqrt{2}x + 1)(\sqrt{2}x - 1)$
(2)$(x + \sqrt{2})^2(x - \sqrt{2})^2$

10. 无论$x$取任何实数，代数式$\sqrt{x^{2}-6x + m}$都有意义，则$m$的取值范围是_______.

$m\geq9$ 解析：$\sqrt{x^2 - 6x + m}=\sqrt{x^2 - 6x + 9 + m - 9}=\sqrt{(x - 3)^2 + m - 9}$.
$\because$无论$x$取任何实数，代数式$\sqrt{x^2 - 6x + m}$都有意义，$\therefore (x - 3)^2 + m - 9\geq0$. 又$\because (x - 3)^2\geq0$，$\therefore m - 9\geq0$，解得$m\geq9$.

11. 若$x$、$y$为实数，且$y = \frac{\sqrt{x^{2}-4}+\sqrt{4 - x^{2}}+1}{x + 2}$，求$x + y - xy$的值.

$\because y = \frac{\sqrt{x^2 - 4}+\sqrt{4 - x^2}+1}{x + 2}$，$\therefore x^2 - 4\geq0$，$4 - x^2\geq0$，$x + 2\neq0$，解得$x = 2$，
$\therefore y = \frac{1}{4}$，$\therefore x + y - xy = 2+\frac{1}{4}-2\times\frac{1}{4}=\frac{7}{4}$.

12. 阅读下面的解题过程，并回答问题.
化简：$(\sqrt{1 - 3x})^{2}-|1 - x|$.
解：由$1 - 3x\geq0$，得$x\leq\frac{1}{3}$，$\therefore1 - x＞0$，
$\therefore$原式$=(1 - 3x)-(1 - x)=1 - 3x - 1 + x=-2x$.
按照上面的解法，解决下列问题.
(1)化简：$(\sqrt{2x + 5})^{2}-(\sqrt{2 - x})^{2}+|x - 3|$.
(2)若$x$满足$|2024 - x|+\sqrt{x - 2025}=x$，求$x - 2024^{2}$的值.

(1)$\because 2x + 5\geq0$且$2 - x\geq0$，$\therefore -\frac{5}{2}\leq x\leq2$，$\therefore$原式$=\vert 2x + 5\vert-\vert 2 - x\vert+\vert x - 3\vert = 2x + 5-(2 - x)+(3 - x)=2x + 6$.
(2)$\because x - 2025\geq0$，$\therefore x\geq2025$，$\therefore 2024 - x＜0$，$\therefore$原式$=x - 2024+\sqrt{x - 2025}=x$，$\therefore \sqrt{x - 2025}=2024$. 两边同时平方得，$x - 2025 = 2024^2$，$\therefore x - 2024^2 = 2025$.

13.（1）已知实数$m$、$n$满足$|4 - 2m|+(n - 2)^{2}+\sqrt{(m - 2)n^{2}}=2m - 4$，求$m + n$的值；
(2) 若实数$a$、$b$、$m$满足如下关系式：$\sqrt{3a - b - m - 4}+\sqrt{a + b - 2024}=3\sqrt{2024 - a - b}-2\sqrt{-a + 3b - m}$，求$|m - 2026|$的平方根.

(1)原式可化为$\vert 4 - 2m\vert+4 - 2m+(n - 2)^2+\sqrt{(m - 2)n^2}=0$.
$\because m - 2\geq0$，$\therefore m\geq2$，$\therefore 4 - 2m\leq0$，$\therefore$原式$=(n - 2)^2+\sqrt{(m - 2)n^2}=0$.
$\because (n - 2)^2\geq0$，$\sqrt{(m - 2)n^2}\geq0$，$\therefore \begin{cases}n - 2 = 0 \\ \sqrt{(m - 2)n^2}=0\end{cases}$，即$\begin{cases}n = 2 \\ m = 2\end{cases}$，$\therefore m + n = 2 + 2 = 4$.
(2)根据题意得，$a + b - 2024\geq0$，$2024 - a - b\geq0$，$\therefore a + b = 2024$，$\therefore$原式可化为$\sqrt{3a - b - m - 4}+2\sqrt{-a + 3b - m}=0$，$\therefore \begin{cases}3a - b - m - 4 = 0 \\ -a + 3b - m = 0\end{cases}$，两式相加得$2a + 2b - 2m - 4 = 0$，$\therefore a + b = m + 2$，$\therefore m + 2 = 2024$，解得$m = 2022$，$\therefore \vert m - 2026\vert = 4$，$\therefore \vert m - 2026\vert$的平方根，即为$4$的平方根，为$\pm2$.