Bài 2:
\(\sqrt{\dfrac{7}{32}}=\sqrt{\dfrac{14}{64}}=\dfrac{\sqrt{14}}{8}\)
\(\sqrt{\dfrac{1}{200}}=\sqrt{\dfrac{2}{400}}=\dfrac{\sqrt{2}}{20}\)
\(\sqrt{\dfrac{5}{18}}=\sqrt{\dfrac{10}{36}}=\dfrac{\sqrt{10}}{6}\)
\(\sqrt{\dfrac{11}{128}}=\sqrt{\dfrac{22}{256}}=\dfrac{\sqrt{22}}{16}\)
\(\sqrt{\dfrac{x^2}{5}}=\dfrac{\left|x\right|}{\sqrt{5}}\)
Bài 1:
6) Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
8) Ta có: \(\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
1) Ta có: \(2\sqrt{7}-\sqrt{28}+3\sqrt{112}-2\sqrt{175}\)
\(=2\sqrt{7}-2\sqrt{7}+12\sqrt{7}-10\sqrt{7}\)
\(=2\sqrt{7}\)
