10. (1)最简二次根式$\sqrt[b - a]{3b}$和$\sqrt{2b - a + 2}$是同类二次根式,则$a + 2b$的值为
4
；
答案：10. (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)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$a - b =$
3
。
答案：(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$\begin{cases}a = -\dfrac{2}{3}\\b = \dfrac{1}{3}\end{cases}$ $2a + 1 = -\dfrac{1}{3}＜0$，不符合题意.故答案为3.
11. 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$m≥ n$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
$-5\sqrt{2}$
。
答案：11. $-5\sqrt{2}$ 解析：$\because$当$m ＜ n$时，$m\varPhi n = 2\sqrt{m}+\sqrt{n}$；当$m≥ n$时，$m\varPhi n = 2\sqrt{m}-\sqrt{n}$，$\therefore(3\varPhi2)-(8\varPhi12)=2\sqrt{3}-\sqrt{2}-(2\sqrt{8}+\sqrt{12})=2\sqrt{3}-\sqrt{2}-4\sqrt{2}-2\sqrt{3}=-5\sqrt{2}$.
解析：
$\because 3≥ 2$，$\therefore 3\varPhi 2=2\sqrt{3}-\sqrt{2}$；
$\because 8＜12$，$\therefore 8\varPhi 12=2\sqrt{8}+\sqrt{12}=4\sqrt{2}+2\sqrt{3}$；
$\therefore (3\varPhi 2)-(8\varPhi 12)=(2\sqrt{3}-\sqrt{2})-(4\sqrt{2}+2\sqrt{3})=2\sqrt{3}-\sqrt{2}-4\sqrt{2}-2\sqrt{3}=-5\sqrt{2}$.
12. (1)先化简,再求值:
$6x\sqrt{\dfrac{y}{x}}+\dfrac{3}{y}\sqrt{xy^{3}}-(4y\sqrt{\dfrac{x}{y}}+\sqrt{36xy})$,其中$x=\sqrt{6}+\sqrt{2}$,$y=\sqrt{6}-\sqrt{2}$；
答案：12. (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)已知$4x^{2}+y^{2}-4x - 6y + 10 = 0$,求$(\dfrac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\dfrac{x}{y^{3}}})-(x^{2}\sqrt{\dfrac{1}{x}}-5x\sqrt{\dfrac{y}{x}})$的值。
答案：(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$x=\dfrac{1}{2}$，$y = 3$. $\because(\dfrac{2}{3}x\sqrt{9x}+y^{2}\sqrt{\dfrac{x}{y^{3}}})-(x^{2}\sqrt{\dfrac{1}{x}}-5x\sqrt{\dfrac{y}{x}})=2x\sqrt{x}+\sqrt{xy}-x\sqrt{x}+5\sqrt{xy}=x\sqrt{x}+6\sqrt{xy}$.当$x=\dfrac{1}{2}$，$y = 3$时，原式$=\dfrac{1}{2}×\sqrt{\dfrac{1}{2}}+6×\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{2}}{4}+3\sqrt{6}$.
13. 已知$M=\dfrac{x + y}{\sqrt{x}-\sqrt{y}}-\dfrac{2xy}{x\sqrt{y}-y\sqrt{x}}$,$N=\dfrac{3\sqrt{x}-2\sqrt{y}}{\sqrt{x + y}+\sqrt{y - x}}$.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。
答案：13. 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$M=\dfrac{x + y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{(\sqrt{x}-\sqrt{y})^{2}}{\sqrt{x}-\sqrt{y}}=\sqrt{x}-\sqrt{y}=\sqrt{8}-\sqrt{18}=-\sqrt{2}$. $N=\dfrac{3\sqrt{8}-2\sqrt{18}}{\sqrt{26}+\sqrt{10}}=\dfrac{6\sqrt{2}-6\sqrt{2}}{\sqrt{26}+\sqrt{10}}=0$. $\therefore M ＜ N$，即$N$的值比$M$大.
解析：
乙同学的说法是正确的.理由：由$y=\sqrt{x - 8}+\sqrt{8 - x}+18$，可得$x=8$，$y=18$.
$M=\dfrac{x + y}{\sqrt{x}-\sqrt{y}}-\dfrac{2xy}{x\sqrt{y}-y\sqrt{x}}=\dfrac{x + y}{\sqrt{x}-\sqrt{y}}-\dfrac{2xy}{\sqrt{xy}(\sqrt{x}-\sqrt{y})}=\dfrac{x + y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{(\sqrt{x}-\sqrt{y})^{2}}{\sqrt{x}-\sqrt{y}}=\sqrt{x}-\sqrt{y}=\sqrt{8}-\sqrt{18}=2\sqrt{2}-3\sqrt{2}=-\sqrt{2}$.
$N=\dfrac{3\sqrt{x}-2\sqrt{y}}{\sqrt{x + y}+\sqrt{y - x}}=\dfrac{3\sqrt{8}-2\sqrt{18}}{\sqrt{8 + 18}+\sqrt{18 - 8}}=\dfrac{6\sqrt{2}-6\sqrt{2}}{\sqrt{26}+\sqrt{10}}=\dfrac{0}{\sqrt{26}+\sqrt{10}}=0$.
$\because -\sqrt{2}＜0$，$\therefore M＜N$，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),
$\therefore2a + 1 = 7m^{2}$,$\therefore a=\dfrac{7m^{2}-1}{2}$.
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：14. $\because\sqrt{2a + 3}$与$\sqrt{5}$能合并，$\therefore\sqrt{2a + 3}=m\sqrt{5}$（$m$为正整数），$\therefore2a + 3 = 5m^{2}$，$\therefore a=\dfrac{5m^{2}-3}{2}$.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）.
