1. (烟台中考)下列二次根式中,与$\sqrt{2}$是同类二次根式的是(
A.$\sqrt{4}$
B.$\sqrt{6}$
C.$\sqrt{8}$
D.$\sqrt{12}$
C
)A.$\sqrt{4}$
B.$\sqrt{6}$
C.$\sqrt{8}$
D.$\sqrt{12}$
答案：1. C 解析：A. $\sqrt{4}=2$，不符合题意；B. $\sqrt{6}$无法继续化简，不符合题意；C. $\sqrt{8}=2\sqrt{2}$，和$\sqrt{2}$是同类二次根式，符合题意；D. $\sqrt{12}=2\sqrt{3}$，不符合题意.故选C.
2. 若$a$,$b$为有理数,且$\sqrt{3}+\sqrt{12}+\sqrt{\dfrac{1}{9}}=a+b\sqrt{3}$,则$ab =$(
A.$\dfrac{1}{3}$
B.$1$
C.$2$
D.$3$
B
)A.$\dfrac{1}{3}$
B.$1$
C.$2$
D.$3$
答案：2. B 解析：$\sqrt{3}+\sqrt{12}+\sqrt{\dfrac{1}{9}}=\sqrt{3}+2\sqrt{3}+\dfrac{1}{3}=\dfrac{1}{3}+3\sqrt{3}$，$\therefore a=\dfrac{1}{3}$，$b = 3$，$\therefore ab = 1$.故选B.
3. 如图,将一根铁丝首尾相接可以围成一个长为$\sqrt{18}π$、宽为$\sqrt{8}π$的矩形,若将这根铁丝展开重新首尾相接围成一个圆形,则该圆的面积是(
A.$25π$
B.$50π$
C.$36π$
D.$72π$
B
)A.$25π$
B.$50π$
C.$36π$
D.$72π$
答案：3. B 解析：铁丝的总长为$2×(\sqrt{18}π+\sqrt{8}π)=2×(3\sqrt{2}π+2\sqrt{2}π)=10\sqrt{2}π$.围成圆形后半径$r$满足$2π r = 10\sqrt{2}π$，$\therefore r = 5\sqrt{2}$，$\therefore$圆的面积为$π r^{2}=50π$.故选B.
4. 计算:
(1)(2025·自贡中考)$\sqrt{18}-3\sqrt{2}=$
(2)(哈尔滨中考)$\sqrt{63}-7\sqrt{\dfrac{1}{7}}=$
(3)(常德中考)$\sqrt{\dfrac{9}{2}}-\sqrt{\dfrac{1}{2}}+\sqrt{8}=$
(1)(2025·自贡中考)$\sqrt{18}-3\sqrt{2}=$
0
；(2)(哈尔滨中考)$\sqrt{63}-7\sqrt{\dfrac{1}{7}}=$
$2\sqrt{7}$
；(3)(常德中考)$\sqrt{\dfrac{9}{2}}-\sqrt{\dfrac{1}{2}}+\sqrt{8}=$
$3\sqrt{2}$
。答案：4. (1) 0 解析：原式$=3\sqrt{2}-3\sqrt{2}=0$.
(2) $2\sqrt{7}$ 解析：原式$=3\sqrt{7}-7×\dfrac{\sqrt{7}}{7}=2\sqrt{7}$.
(3) $3\sqrt{2}$ 解析：原式$=\dfrac{3\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}+2\sqrt{2}=3\sqrt{2}$.
(2) $2\sqrt{7}$ 解析：原式$=3\sqrt{7}-7×\dfrac{\sqrt{7}}{7}=2\sqrt{7}$.
(3) $3\sqrt{2}$ 解析：原式$=\dfrac{3\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}+2\sqrt{2}=3\sqrt{2}$.
5. 若$\sqrt{18x}+2\sqrt{\dfrac{x}{2}}+x\sqrt{\dfrac{2}{x}}=10$,则$x$的值为
2
。答案：5. 2 解析：$\sqrt{18x}+2\sqrt{\dfrac{x}{2}}+x\sqrt{\dfrac{2}{x}}=3\sqrt{2x}+2×\dfrac{\sqrt{2x}}{2}+x\dfrac{\sqrt{2x}}{x}=5\sqrt{2x}$，$\therefore5\sqrt{2x}=10$，$\therefore x = 2$.
解析：
解：$\sqrt{18x}+2\sqrt{\dfrac{x}{2}}+x\sqrt{\dfrac{2}{x}}$
$=3\sqrt{2x}+2×\dfrac{\sqrt{2x}}{2}+x×\dfrac{\sqrt{2x}}{x}$
$=3\sqrt{2x}+\sqrt{2x}+\sqrt{2x}$
$=5\sqrt{2x}$
因为$5\sqrt{2x}=10$，所以$\sqrt{2x}=2$，两边平方得$2x=4$，解得$x=2$。
2
$=3\sqrt{2x}+2×\dfrac{\sqrt{2x}}{2}+x×\dfrac{\sqrt{2x}}{x}$
$=3\sqrt{2x}+\sqrt{2x}+\sqrt{2x}$
$=5\sqrt{2x}$
因为$5\sqrt{2x}=10$，所以$\sqrt{2x}=2$，两边平方得$2x=4$，解得$x=2$。
2
6. 教材变式 计算:
(1)$2\sqrt{12}-6\sqrt{\dfrac{1}{3}}+\sqrt{27}$；
(2)$\sqrt{27}-4\sqrt{\dfrac{1}{8}}-(\sqrt{3}-\sqrt{8})$；
(3)$\dfrac{2}{3}\sqrt{9x}+6\sqrt{\dfrac{x}{4}}-x\sqrt{\dfrac{1}{x}}(x＞0)$；
(4)$4b\sqrt{\dfrac{a}{b}}-(\dfrac{2}{a}\sqrt{a^{3}b}+\sqrt{25ab})(a＞0,b＞0)$。
(1)$2\sqrt{12}-6\sqrt{\dfrac{1}{3}}+\sqrt{27}$；
(2)$\sqrt{27}-4\sqrt{\dfrac{1}{8}}-(\sqrt{3}-\sqrt{8})$；
(3)$\dfrac{2}{3}\sqrt{9x}+6\sqrt{\dfrac{x}{4}}-x\sqrt{\dfrac{1}{x}}(x＞0)$；
(4)$4b\sqrt{\dfrac{a}{b}}-(\dfrac{2}{a}\sqrt{a^{3}b}+\sqrt{25ab})(a＞0,b＞0)$。
答案：6. (1) 原式$=4\sqrt{3}-6×\dfrac{\sqrt{3}}{3}+3\sqrt{3}=5\sqrt{3}$.
(2) 原式$=3\sqrt{3}-\sqrt{2}-\sqrt{3}+2\sqrt{2}=2\sqrt{3}+\sqrt{2}$.
(3) 原式$=\dfrac{2}{3}·3\sqrt{x}+6·\dfrac{\sqrt{x}}{2}-x·\dfrac{\sqrt{x}}{x}=2\sqrt{x}+3\sqrt{x}-\sqrt{x}=4\sqrt{x}$.
(4) 原式$=4b·\dfrac{\sqrt{ab}}{b}-\dfrac{2}{a}· a\sqrt{ab}-5\sqrt{ab}=4\sqrt{ab}-2\sqrt{ab}-5\sqrt{ab}=-3\sqrt{ab}$.
(2) 原式$=3\sqrt{3}-\sqrt{2}-\sqrt{3}+2\sqrt{2}=2\sqrt{3}+\sqrt{2}$.
(3) 原式$=\dfrac{2}{3}·3\sqrt{x}+6·\dfrac{\sqrt{x}}{2}-x·\dfrac{\sqrt{x}}{x}=2\sqrt{x}+3\sqrt{x}-\sqrt{x}=4\sqrt{x}$.
(4) 原式$=4b·\dfrac{\sqrt{ab}}{b}-\dfrac{2}{a}· a\sqrt{ab}-5\sqrt{ab}=4\sqrt{ab}-2\sqrt{ab}-5\sqrt{ab}=-3\sqrt{ab}$.
7. 化简$\sqrt{-x^{3}}-x\sqrt{-\dfrac{1}{x}}$,得(
A.$(x - 1)\sqrt{-x}$
B.$(1 - x)\sqrt{-x}$
C.$-(x + 1)\sqrt{x}$
D.$(x - 1)\sqrt{x}$
B
)A.$(x - 1)\sqrt{-x}$
B.$(1 - x)\sqrt{-x}$
C.$-(x + 1)\sqrt{x}$
D.$(x - 1)\sqrt{x}$
答案：7. B 解析：$\because$根据二次根式的性质知，$-x＞0$，即$x＜0$，$\therefore\sqrt{-x^{3}}-x\sqrt{-\dfrac{1}{x}}=\vert x\vert\sqrt{-x}-\dfrac{x}{\vert x\vert}\sqrt{-x}=-x\sqrt{-x}+\sqrt{-x}=(1 - x)\sqrt{-x}$.故选B.
8. 已知整数$x$,$y$满足$\sqrt{x}+2\sqrt{y}=\sqrt{50}$,那么整数对$(x,y)$的个数是(
A.$0$
B.$1$
C.$2$
D.$3$
D
)A.$0$
B.$1$
C.$2$
D.$3$
答案：8. D 解析：$\because\sqrt{50}=5\sqrt{2}+0=\sqrt{50}+2\sqrt{0}$，$\therefore x = 50$，$y = 0$. $\because\sqrt{50}=5\sqrt{2}=\sqrt{2}+4\sqrt{2}=\sqrt{2}+2\sqrt{8}$，$\therefore x = 2$，$y = 8$. $\because\sqrt{50}=5\sqrt{2}=3\sqrt{2}+2\sqrt{2}=\sqrt{18}+2\sqrt{2}$，$\therefore x = 18$，$y = 2$. 综上，满足条件的整数对$(x,y)$有3个：$(50,0)$，$(2,8)$，$(18,2)$.故选D.
9. 一个等腰三角形的两边长分别为$8\sqrt{\dfrac{1}{8}}$,$12\sqrt{\dfrac{1}{2}}$,则这个等腰三角形的周长为
$14\sqrt{2}$
。答案：9. $14\sqrt{2}$ 解析：$8\sqrt{\dfrac{1}{8}}=2\sqrt{2}$，$12\sqrt{\dfrac{1}{2}}=6\sqrt{2}$.有两种可能：①腰长为$2\sqrt{2}$，底边长为$6\sqrt{2}$，$\because2\sqrt{2}+2\sqrt{2}＜6\sqrt{2}$，$\therefore$此时不能构成三角形；②腰长为$6\sqrt{2}$，底边长为$2\sqrt{2}$，$\because2\sqrt{2}+6\sqrt{2}＞6\sqrt{2}$，$\therefore$此时能构成三角形，其周长为$6\sqrt{2}+6\sqrt{2}+2\sqrt{2}=14\sqrt{2}$.故这个等腰三角形的周长为$14\sqrt{2}$.