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![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng quy tắc khai phương một tích
1: Ta có: \(\sqrt{\frac{1}{5}}\cdot\sqrt{\frac{1}{20}}\cdot3\cdot7\)
\(=\sqrt{\frac{1}{5}}\cdot\sqrt{\frac{1}{20}}\cdot\sqrt{9}\cdot\sqrt{49}\)
\(=\sqrt{\frac{1}{5}\cdot\frac{1}{20}\cdot9\cdot49}\)
\(=\sqrt{\frac{441}{100}}=\frac{\sqrt{441}}{\sqrt{100}}=\frac{21}{10}\)
2: Ta có: \(\sqrt{0,001\cdot360\cdot3^2\cdot\left(-3\right)^2}\)
\(=\sqrt{0,001}\cdot\sqrt{360}\cdot\sqrt{3^{^2}}\cdot\sqrt{\left(-3\right)^2}\)
\(=\sqrt{\frac{1}{100}}\cdot\sqrt{\frac{1}{10}}\cdot\sqrt{6^2}\cdot\sqrt{10}\cdot3\cdot3\)
\(=\frac{1}{10}\cdot6\cdot9\cdot\sqrt{\frac{1}{10}\cdot10}=\frac{54}{10}\cdot1=\frac{27}{5}\)
Áp dụng quy tắc nhân căn thức bậc hai
1: Ta có: \(2\sqrt{2}\left(4\sqrt{8}-\sqrt{32}\right)\)
\(=2\sqrt{2}\cdot4\sqrt{8}-2\sqrt{2}\cdot\sqrt{32}\)
\(=8\cdot\sqrt{16}-2\cdot\sqrt{64}\)
\(=8\cdot4-2\cdot8\)
=32-16=16
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)
=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)
=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)
=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}\)
b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
em hổng có biết đâu vì em chưa hc lp 9 mới lại đề bài dài kinh khủng