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10. (1) 最简二次根式$\sqrt[a - b]{3b}$和$\sqrt{2b - a + 2}$是同类二次根式，则$a + 2b$的值为_______；
(2) 最简二次根式$\sqrt[|a - b - 1|]{2a + 1}$和$\sqrt{3a + 2b + 1}$是同类二次根式，则$a - b =$_______.
(2) 最简二次根式$\sqrt[|a - b - 1|]{2a + 1}$和$\sqrt{3a + 2b + 1}$是同类二次根式，则$a - b =$_______.
答案
(1) 4 解析：由题意得$\begin{cases}b - a = 2\\3b = 2b - a + 2\end{cases}$，解得$\begin{cases}b = 2\\a = 0\end{cases}$，$\therefore a + 2b = 4$.
(2) 3 解析：根据题意，得$\begin{cases}\vert a - b - 1\vert = 2\\2a + 1 = 3a + 2b + 1\end{cases}$. 当$a - b - 1＞0$时，解得$\begin{cases}a = 2\\b = -1\end{cases}$，代入原式结果均为$\sqrt{5}$，符合条件，此时$a - b = 3$；当$a - b - 1＜0$时，解得$\begin{cases}a = -\frac{2}{3}\\b = \frac{1}{3}\end{cases}$，$2a + 1 = -\frac{1}{3}＜0$，不符合题意. 故答案为 3.
(2) 3 解析：根据题意，得$\begin{cases}\vert a - b - 1\vert = 2\\2a + 1 = 3a + 2b + 1\end{cases}$. 当$a - b - 1＞0$时，解得$\begin{cases}a = 2\\b = -1\end{cases}$，代入原式结果均为$\sqrt{5}$，符合条件，此时$a - b = 3$；当$a - b - 1＜0$时，解得$\begin{cases}a = -\frac{2}{3}\\b = \frac{1}{3}\end{cases}$，$2a + 1 = -\frac{1}{3}＜0$，不符合题意. 故答案为 3.
11.（2024·齐齐哈尔期中）我们规定：对于任意的正数$m、n$的运算“$\varPhi$”为：当$m ＜ n$时，$m\varPhi n = 2\sqrt{m}+\sqrt{n}$；当$m\geq n$时，$m\varPhi n = 2\sqrt{m}-\sqrt{n}$，其他运算符号意义不变. 按上述规定，计算$(3\varPhi2)-(8\varPhi12)$的结果为_______.
答案
$-5\sqrt{2}$ 解析：$\because$当$m ＜ n$时，$m\varPhi n = 2\sqrt{m}+\sqrt{n}$；当$m\geq n$时，$m\varPhi n = 2\sqrt{m}-\sqrt{n}$，$\therefore(3\varPhi2)-(8\varPhi12)=2\sqrt{3}-\sqrt{2}-(2\sqrt{8}+\sqrt{12})=2\sqrt{3}-\sqrt{2}-4\sqrt{2}-2\sqrt{3}=-5\sqrt{2}$.
12. (1) 先化简，再求值：
$6x\sqrt{\frac{y}{x}}+\frac{3}{y}\sqrt{xy^{3}}-(4y\sqrt{\frac{x}{y}}+\sqrt{36xy})$，其中$x = \sqrt{6}+\sqrt{2},y = \sqrt{6}-\sqrt{2}$；
(2) 已知$4x^{2}+y^{2}-4x - 6y + 10 = 0$，求$(\frac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\frac{x}{y^{3}}})-(x^{2}\sqrt{\frac{1}{x}}-5x\sqrt{\frac{y}{x}})$的值.
$6x\sqrt{\frac{y}{x}}+\frac{3}{y}\sqrt{xy^{3}}-(4y\sqrt{\frac{x}{y}}+\sqrt{36xy})$，其中$x = \sqrt{6}+\sqrt{2},y = \sqrt{6}-\sqrt{2}$；
(2) 已知$4x^{2}+y^{2}-4x - 6y + 10 = 0$，求$(\frac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\frac{x}{y^{3}}})-(x^{2}\sqrt{\frac{1}{x}}-5x\sqrt{\frac{y}{x}})$的值.
答案
(1)原式$=6\sqrt{xy}+3\sqrt{xy}-4\sqrt{xy}-6\sqrt{xy}=-\sqrt{xy}$.$\because x = \sqrt{6}+\sqrt{2}$，$y = \sqrt{6}-\sqrt{2}$，$\therefore$原式$=-\sqrt{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=-\sqrt{6 - 2}=-\sqrt{4}=-2$.
(2)$\because4x^{2}+y^{2}-4x - 6y + 10 = 0$，$\therefore(2x - 1)^{2}+(y - 3)^{2}=0$，$\therefore2x - 1 = 0$，$y - 3 = 0$，解得$x = \frac{1}{2}$，$y = 3$.$\because(\frac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\frac{x}{y^{3}}})-(x^{2}\sqrt{\frac{1}{x}}-5x\sqrt{\frac{y}{x}})=2x\sqrt{x}+ \sqrt{xy}-x\sqrt{x}+5\sqrt{xy}=x\sqrt{x}+6\sqrt{xy}$. 当$x = \frac{1}{2}$，$y = 3$时，原式$=\frac{1}{2}\times\sqrt{\frac{1}{2}}+6\times\sqrt{\frac{3}{2}}=\frac{\sqrt{2}}{4}+3\sqrt{6}$.
(2)$\because4x^{2}+y^{2}-4x - 6y + 10 = 0$，$\therefore(2x - 1)^{2}+(y - 3)^{2}=0$，$\therefore2x - 1 = 0$，$y - 3 = 0$，解得$x = \frac{1}{2}$，$y = 3$.$\because(\frac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\frac{x}{y^{3}}})-(x^{2}\sqrt{\frac{1}{x}}-5x\sqrt{\frac{y}{x}})=2x\sqrt{x}+ \sqrt{xy}-x\sqrt{x}+5\sqrt{xy}=x\sqrt{x}+6\sqrt{xy}$. 当$x = \frac{1}{2}$，$y = 3$时，原式$=\frac{1}{2}\times\sqrt{\frac{1}{2}}+6\times\sqrt{\frac{3}{2}}=\frac{\sqrt{2}}{4}+3\sqrt{6}$.
13. 已知$M=\frac{x + y}{\sqrt{x}-\sqrt{y}}-\frac{2xy}{x\sqrt{y}-y\sqrt{x}},N=\frac{3\sqrt{x}-2\sqrt{y}}{\sqrt{x + y}+\sqrt{y - x}}$.
甲、乙两个同学在$y=\sqrt{x - 8}+\sqrt{8 - x}+18$的条件下分别计算了$M$和$N$的值. 甲说$M$的值比$N$大，乙说$N$的值比$M$大. 请你判断他们谁的说法是正确的，并说明理由.
甲、乙两个同学在$y=\sqrt{x - 8}+\sqrt{8 - x}+18$的条件下分别计算了$M$和$N$的值. 甲说$M$的值比$N$大，乙说$N$的值比$M$大. 请你判断他们谁的说法是正确的，并说明理由.
答案
乙同学的说法是正确的. 理由：由$y = \sqrt{x - 8}+\sqrt{8 - x}+18$，可得$x = 8$，$y = 18$. 因此$M=\frac{x + y - 2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\frac{(\sqrt{x}-\sqrt{y})^{2}}{\sqrt{x}-\sqrt{y}}=\sqrt{x}-\sqrt{y}=\sqrt{8}-\sqrt{18}=-\sqrt{2}$. $N=\frac{3\sqrt{8}-2\sqrt{18}}{\sqrt{26}+\sqrt{10}}=\frac{6\sqrt{2}-6\sqrt{2}}{\sqrt{26}+\sqrt{10}}=0$.$\therefore M ＜ N$，即$N$的值比$M$大.
14. 已知$a$为正整数，且$\sqrt{2a + 1}$与$\sqrt{7}$能合并，试写出三个满足条件的$a$的值.
解：$\because\sqrt{2a + 1}$与$\sqrt{7}$能合并，
$\therefore\sqrt{2a + 1}=m\sqrt{7}(m$为正整数$)$，
$\therefore2a + 1 = 7m^{2},\therefore a=\frac{7m^{2}-1}{2}$.
又$a$为正整数，$\therefore7m^{2}-1$为偶数，$\therefore m$为正奇数，
$\therefore$当$m = 1$时，$a = 3$；当$m = 3$时，$a = 31$；当$m = 5$时，$a = 87$，
$\therefore$满足条件的$a$的值可以为$3,31,87$（也可取$m$为其他正奇数，得出不同的答案）.
请根据上面的信息，回答问题：
已知$a$为正整数，且$\sqrt{2a + 3}$与$\sqrt{5}$能合并，试写出三个满足条件的$a$的值.
解：$\because\sqrt{2a + 1}$与$\sqrt{7}$能合并，
$\therefore\sqrt{2a + 1}=m\sqrt{7}(m$为正整数$)$，
$\therefore2a + 1 = 7m^{2},\therefore a=\frac{7m^{2}-1}{2}$.
又$a$为正整数，$\therefore7m^{2}-1$为偶数，$\therefore m$为正奇数，
$\therefore$当$m = 1$时，$a = 3$；当$m = 3$时，$a = 31$；当$m = 5$时，$a = 87$，
$\therefore$满足条件的$a$的值可以为$3,31,87$（也可取$m$为其他正奇数，得出不同的答案）.
请根据上面的信息，回答问题：
已知$a$为正整数，且$\sqrt{2a + 3}$与$\sqrt{5}$能合并，试写出三个满足条件的$a$的值.
答案
$\because\sqrt{2a + 3}$与$\sqrt{5}$能合并，$\therefore\sqrt{2a + 3}=m\sqrt{5}$（$m$为正整数），$\therefore2a + 3 = 5m^{2}$，$\therefore a=\frac{5m^{2}-3}{2}$. 又$a$为正整数，$\therefore5m^{2}-3$为偶数，$\therefore m$为正奇数，$\therefore$当$m = 1$时，$a = 1$；当$m = 3$时，$a = 21$；当$m = 5$时，$a = 61$，$\therefore$满足条件的$a$的值可以为 1，21，61（也可取$m$为其他正奇数，得出不同的答案）.
