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一、填空题
1. $\sqrt{15} × \sqrt{5} - \sqrt{3} =$
2. $( \sqrt{27} + \sqrt{\dfrac{4}{3}} ) × \sqrt{3} =$
3. $( \sqrt{24} - \sqrt{6} ) ÷ \sqrt{3} =$
4. $( \sqrt{3} + 1 ) ( 1 - \sqrt{3} ) =$
5. $\dfrac{\sqrt{2}}{3} × ( \sqrt{8} + \sqrt{2} ) =$
6. $\sqrt{2} × ( \sqrt{2} + \sqrt{3} ) - \sqrt{24} =$
1. $\sqrt{15} × \sqrt{5} - \sqrt{3} =$
$4\sqrt{3}$
2. $( \sqrt{27} + \sqrt{\dfrac{4}{3}} ) × \sqrt{3} =$
11
3. $( \sqrt{24} - \sqrt{6} ) ÷ \sqrt{3} =$
$\sqrt{2}$
4. $( \sqrt{3} + 1 ) ( 1 - \sqrt{3} ) =$
-2
5. $\dfrac{\sqrt{2}}{3} × ( \sqrt{8} + \sqrt{2} ) =$
2
6. $\sqrt{2} × ( \sqrt{2} + \sqrt{3} ) - \sqrt{24} =$
$2-\sqrt{6}$
答案
1. $4\sqrt{3}$
2. 11
3. $\sqrt{2}$
4. -2
5. 2
6. $2-\sqrt{6}$
2. 11
3. $\sqrt{2}$
4. -2
5. 2
6. $2-\sqrt{6}$
7. $(\sqrt{\frac{1}{2}}+\sqrt{3})×(\sqrt{2}-\sqrt{12})=$
8. $(\sqrt{2}+1)(\sqrt{2}-1)^2=$
-5
8. $(\sqrt{2}+1)(\sqrt{2}-1)^2=$
$\sqrt{2}-1$
答案
7. -5
8. $\sqrt{2}-1$
8. $\sqrt{2}-1$
二、解答题
9. $(\sqrt{24} - \sqrt{\frac{1}{2}}) - 2(\sqrt{\frac{1}{8}} - \sqrt{6})$
10. $(3+\sqrt{5})(3-\sqrt{5}) - (\sqrt{3}-1)^2$
11. $(\frac{1}{2}\sqrt{28} - \frac{3}{2}\sqrt{84}) × \sqrt{14}$
12. $\sqrt{27} + (4\sqrt{2} - 2\sqrt{6}) ÷ 2\sqrt{2}$
9. $(\sqrt{24} - \sqrt{\frac{1}{2}}) - 2(\sqrt{\frac{1}{8}} - \sqrt{6})$
10. $(3+\sqrt{5})(3-\sqrt{5}) - (\sqrt{3}-1)^2$
11. $(\frac{1}{2}\sqrt{28} - \frac{3}{2}\sqrt{84}) × \sqrt{14}$
12. $\sqrt{27} + (4\sqrt{2} - 2\sqrt{6}) ÷ 2\sqrt{2}$
答案
9. $4\sqrt{6}-\sqrt{2}$
10. $2\sqrt{3}$
11. $7\sqrt{2}-21\sqrt{6}$
12. $2\sqrt{3}+2$
10. $2\sqrt{3}$
11. $7\sqrt{2}-21\sqrt{6}$
12. $2\sqrt{3}+2$
13. 易错题 先化简,再求值:$(\sqrt{\dfrac{1}{a}}-\sqrt{b})×\sqrt{ab}$,其中$a=2,b=8.$
答案
13. 原式=$\sqrt{b}-b\sqrt{a}$. 当$a=2,b=8$时,原式=$-6\sqrt{2}$
易错分析：13. 易在运用乘法对加法的分配律展开或代入求值时出现计算错误.
易错分析：13. 易在运用乘法对加法的分配律展开或代入求值时出现计算错误.
14. 已知$a=3+\sqrt{2},b=3-\sqrt{2}$,求代数式$\sqrt{a^2 -3ab +b^2}$的值.
答案
14. 原式=$\sqrt{(a-b)^2 -ab}$. 当$a=3+\sqrt{2},b=3-\sqrt{2}$时,原式=$\sqrt{(2\sqrt{2})^2 -7}=1$
