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17. 计算下列各题.
(1)$\sqrt{7} + 2\sqrt{7} + 3\sqrt{9 × 7}$；
(2)$3\sqrt{2} + \sqrt{3} - 2\sqrt{2} - 3\sqrt{3}$；
(3)$\sqrt{8} + \sqrt{18}$；
(4)$2\sqrt{3} - \sqrt{8} + \frac{1}{2}\sqrt{12} + \frac{1}{5}\sqrt{50}$；
(5)$\sqrt{24} ÷ \sqrt{3}$；
(6)$\sqrt{28} × \sqrt{33} ÷ \sqrt{21}$；
(7)$\sqrt{16} × \sqrt{\frac{3}{4}} ÷ \sqrt{3}$；
(8)$\frac{\sqrt{27} - \sqrt{12}}{\sqrt{3}}$.
(1)$\sqrt{7} + 2\sqrt{7} + 3\sqrt{9 × 7}$；
(2)$3\sqrt{2} + \sqrt{3} - 2\sqrt{2} - 3\sqrt{3}$；
(3)$\sqrt{8} + \sqrt{18}$；
(4)$2\sqrt{3} - \sqrt{8} + \frac{1}{2}\sqrt{12} + \frac{1}{5}\sqrt{50}$；
(5)$\sqrt{24} ÷ \sqrt{3}$；
(6)$\sqrt{28} × \sqrt{33} ÷ \sqrt{21}$；
(7)$\sqrt{16} × \sqrt{\frac{3}{4}} ÷ \sqrt{3}$；
(8)$\frac{\sqrt{27} - \sqrt{12}}{\sqrt{3}}$.
答案
解：
(1)
$\begin{aligned}\mathrm{原式}&=\sqrt{7}+2\sqrt{7}+3×\sqrt{9}×\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+3×3\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+9\sqrt{7}\\&=(1+2+9)\sqrt{7}\\&=12\sqrt{7}\end{aligned}$
(2)
$\begin{aligned}\mathrm{原式}&=(3\sqrt{2}-2\sqrt{2})+(\sqrt{3}-3\sqrt{3})\\&=\sqrt{2}-2\sqrt{3}\end{aligned}$
(3)
$\begin{aligned}\mathrm{原式}&=2\sqrt{2}+3\sqrt{2}\\&=5\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}\mathrm{原式}&=2\sqrt{3}-2\sqrt{2}+\frac{1}{2}×2\sqrt{3}+\frac{1}{5}×5\sqrt{2}\\&=2\sqrt{3}-2\sqrt{2}+\sqrt{3}+\sqrt{2}\\&=(2\sqrt{3}+\sqrt{3})+(-2\sqrt{2}+\sqrt{2})\\&=3\sqrt{3}-\sqrt{2}\end{aligned}$
(5)
$\begin{aligned}\mathrm{原式}&=\sqrt{24÷3}\\&=\sqrt{8}\\&=2\sqrt{2}\end{aligned}$
(6)
$\begin{aligned}\mathrm{原式}&=\sqrt{28×33÷21}\\&=\sqrt{44}\\&=2\sqrt{11}\end{aligned}$
(7)
$\begin{aligned}\mathrm{原式}&=4×\frac{\sqrt{3}}{2}÷\sqrt{3}\\&=2\sqrt{3}÷\sqrt{3}\\&=2\end{aligned}$
(8)
$\begin{aligned}\mathrm{原式}&=\frac{\sqrt{27}}{\sqrt{3}}-\frac{\sqrt{12}}{\sqrt{3}}\\&=\sqrt{27÷3}-\sqrt{12÷3}\\&=\sqrt{9}-\sqrt{4}\\&=3-2\\&=1\end{aligned}$
(1)
$\begin{aligned}\mathrm{原式}&=\sqrt{7}+2\sqrt{7}+3×\sqrt{9}×\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+3×3\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+9\sqrt{7}\\&=(1+2+9)\sqrt{7}\\&=12\sqrt{7}\end{aligned}$
(2)
$\begin{aligned}\mathrm{原式}&=(3\sqrt{2}-2\sqrt{2})+(\sqrt{3}-3\sqrt{3})\\&=\sqrt{2}-2\sqrt{3}\end{aligned}$
(3)
$\begin{aligned}\mathrm{原式}&=2\sqrt{2}+3\sqrt{2}\\&=5\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}\mathrm{原式}&=2\sqrt{3}-2\sqrt{2}+\frac{1}{2}×2\sqrt{3}+\frac{1}{5}×5\sqrt{2}\\&=2\sqrt{3}-2\sqrt{2}+\sqrt{3}+\sqrt{2}\\&=(2\sqrt{3}+\sqrt{3})+(-2\sqrt{2}+\sqrt{2})\\&=3\sqrt{3}-\sqrt{2}\end{aligned}$
(5)
$\begin{aligned}\mathrm{原式}&=\sqrt{24÷3}\\&=\sqrt{8}\\&=2\sqrt{2}\end{aligned}$
(6)
$\begin{aligned}\mathrm{原式}&=\sqrt{28×33÷21}\\&=\sqrt{44}\\&=2\sqrt{11}\end{aligned}$
(7)
$\begin{aligned}\mathrm{原式}&=4×\frac{\sqrt{3}}{2}÷\sqrt{3}\\&=2\sqrt{3}÷\sqrt{3}\\&=2\end{aligned}$
(8)
$\begin{aligned}\mathrm{原式}&=\frac{\sqrt{27}}{\sqrt{3}}-\frac{\sqrt{12}}{\sqrt{3}}\\&=\sqrt{27÷3}-\sqrt{12÷3}\\&=\sqrt{9}-\sqrt{4}\\&=3-2\\&=1\end{aligned}$
解析
【分析】
这组题是二次根式的基础运算题，解题思路可按运算类型拆分：①二次根式乘除运算：利用二次根式乘除法则$\sqrt{a} · \sqrt{b}=\sqrt{ab}(a≥0,b≥0)$、$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}(a≥0,b＞0)$，先把运算统一到同一根号内计算，再将结果化为最简二次根式；②二次根式加减运算：先把所有二次根式化为最简二次根式，再合并同类二次根式（仅合并系数，被开方数和根号保持不变）；③混合运算遵循整式的运算顺序和运算律，可利用分配律等简化计算。
【解析】
(1)
$\begin{aligned}\mathrm{原式}&=\sqrt{7}+2\sqrt{7}+3×\sqrt{9}×\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+3×3\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+9\sqrt{7}\\&=(1+2+9)\sqrt{7}\\&=12\sqrt{7}\end{aligned}$
(2)
$\begin{aligned}\mathrm{原式}&=(3\sqrt{2}-2\sqrt{2})+(\sqrt{3}-3\sqrt{3})\\&=\sqrt{2}-2\sqrt{3}\end{aligned}$
(3)
$\begin{aligned}\mathrm{原式}&=2\sqrt{2}+3\sqrt{2}\\&=5\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}\mathrm{原式}&=2\sqrt{3}-2\sqrt{2}+\frac{1}{2}×2\sqrt{3}+\frac{1}{5}×5\sqrt{2}\\&=2\sqrt{3}-2\sqrt{2}+\sqrt{3}+\sqrt{2}\\&=(2\sqrt{3}+\sqrt{3})+(-2\sqrt{2}+\sqrt{2})\\&=3\sqrt{3}-\sqrt{2}\end{aligned}$
(5)
$\begin{aligned}\mathrm{原式}&=\sqrt{24÷3}\\&=\sqrt{8}\\&=2\sqrt{2}\end{aligned}$
(6)
$\begin{aligned}\mathrm{原式}&=\sqrt{28×33÷21}\\&=\sqrt{44}\\&=2\sqrt{11}\end{aligned}$
(7)
$\begin{aligned}\mathrm{原式}&=4×\frac{\sqrt{3}}{2}÷\sqrt{3}\\&=2\sqrt{3}÷\sqrt{3}\\&=2\end{aligned}$
(8)
$\begin{aligned}\mathrm{原式}&=\frac{\sqrt{27}}{\sqrt{3}}-\frac{\sqrt{12}}{\sqrt{3}}\\&=\sqrt{27÷3}-\sqrt{12÷3}\\&=\sqrt{9}-\sqrt{4}\\&=3-2\\&=1\end{aligned}$
【答案】
(1)$12\sqrt{7}$；(2)$\sqrt{2}-2\sqrt{3}$；(3)$5\sqrt{2}$；(4)$3\sqrt{3}-\sqrt{2}$；(5)$2\sqrt{2}$；(6)$2\sqrt{11}$；(7)$2$；(8)$1$
【知识点】
二次根式化简，同类二次根式合并，二次根式乘除运算
【点评】
本组题是二次根式运算的常规基础题，核心考察二次根式的基本运算法则和化简能力，熟练掌握最简二次根式的化简方法、同类二次根式的合并规则是解题的关键，通过练习可有效提升运算的速度和准确率。
【难度系数】
0.8
这组题是二次根式的基础运算题，解题思路可按运算类型拆分：①二次根式乘除运算：利用二次根式乘除法则$\sqrt{a} · \sqrt{b}=\sqrt{ab}(a≥0,b≥0)$、$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}(a≥0,b＞0)$，先把运算统一到同一根号内计算，再将结果化为最简二次根式；②二次根式加减运算：先把所有二次根式化为最简二次根式，再合并同类二次根式（仅合并系数，被开方数和根号保持不变）；③混合运算遵循整式的运算顺序和运算律，可利用分配律等简化计算。
【解析】
(1)
$\begin{aligned}\mathrm{原式}&=\sqrt{7}+2\sqrt{7}+3×\sqrt{9}×\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+3×3\sqrt{7}\\&=\sqrt{7}+2\sqrt{7}+9\sqrt{7}\\&=(1+2+9)\sqrt{7}\\&=12\sqrt{7}\end{aligned}$
(2)
$\begin{aligned}\mathrm{原式}&=(3\sqrt{2}-2\sqrt{2})+(\sqrt{3}-3\sqrt{3})\\&=\sqrt{2}-2\sqrt{3}\end{aligned}$
(3)
$\begin{aligned}\mathrm{原式}&=2\sqrt{2}+3\sqrt{2}\\&=5\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}\mathrm{原式}&=2\sqrt{3}-2\sqrt{2}+\frac{1}{2}×2\sqrt{3}+\frac{1}{5}×5\sqrt{2}\\&=2\sqrt{3}-2\sqrt{2}+\sqrt{3}+\sqrt{2}\\&=(2\sqrt{3}+\sqrt{3})+(-2\sqrt{2}+\sqrt{2})\\&=3\sqrt{3}-\sqrt{2}\end{aligned}$
(5)
$\begin{aligned}\mathrm{原式}&=\sqrt{24÷3}\\&=\sqrt{8}\\&=2\sqrt{2}\end{aligned}$
(6)
$\begin{aligned}\mathrm{原式}&=\sqrt{28×33÷21}\\&=\sqrt{44}\\&=2\sqrt{11}\end{aligned}$
(7)
$\begin{aligned}\mathrm{原式}&=4×\frac{\sqrt{3}}{2}÷\sqrt{3}\\&=2\sqrt{3}÷\sqrt{3}\\&=2\end{aligned}$
(8)
$\begin{aligned}\mathrm{原式}&=\frac{\sqrt{27}}{\sqrt{3}}-\frac{\sqrt{12}}{\sqrt{3}}\\&=\sqrt{27÷3}-\sqrt{12÷3}\\&=\sqrt{9}-\sqrt{4}\\&=3-2\\&=1\end{aligned}$
【答案】
(1)$12\sqrt{7}$；(2)$\sqrt{2}-2\sqrt{3}$；(3)$5\sqrt{2}$；(4)$3\sqrt{3}-\sqrt{2}$；(5)$2\sqrt{2}$；(6)$2\sqrt{11}$；(7)$2$；(8)$1$
【知识点】
二次根式化简，同类二次根式合并，二次根式乘除运算
【点评】
本组题是二次根式运算的常规基础题，核心考察二次根式的基本运算法则和化简能力，熟练掌握最简二次根式的化简方法、同类二次根式的合并规则是解题的关键，通过练习可有效提升运算的速度和准确率。
【难度系数】
0.8
