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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)
b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(-2\sqrt{5}+\sqrt{3}-6\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)
\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{15}{\sqrt{6}-1}+\dfrac{8}{\sqrt{6}+2}+\dfrac{6}{3-\sqrt{6}}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{8\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\dfrac{6\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{6-1}+\dfrac{8\left(\sqrt{6}-2\right)}{6-4}+\dfrac{6\left(3+\sqrt{6}\right)}{9-6}-9\sqrt{6}\)
\(=3\left(\sqrt{6}+1\right)+4\left(\sqrt{6}-2\right)+2\left(3+\sqrt{6}\right)-9\sqrt{6}\)
\(=3\sqrt{6}+3+4\sqrt{6}-8+6+2\sqrt{6}-9\sqrt{6}\)
\(=9\sqrt{6}+1-9\sqrt{6}\)
\(=1\)
\(\sqrt{\left(\sqrt{5}-1\right)\sqrt{14-6\sqrt{5}}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{9-6\sqrt{5}+5}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left|3-\sqrt{5}\right|}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{3\sqrt{5}-5-3+\sqrt{5}}\)
\(=\sqrt{4\sqrt{5}-8}\)
\(=\sqrt{4\left(\sqrt{5}-2\right)}\)
\(=2\sqrt{\sqrt{5}-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)
\(=\dfrac{-2\sqrt{6}\left(5-\sqrt{6}\right)}{5-\sqrt{6}}-\sqrt{\dfrac{2}{3}\cdot9}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)
\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)
\(=-3\sqrt{6}+3\sqrt{6}+3\)
=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}\)
\(=0\)
c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)
\(=\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
\(=4-4\sqrt{5}+5\)
\(=9-4\sqrt{5}\)
d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121\)
\(=-115\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}=\frac{3\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)
2.
\(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)
3.
\(\frac{7+4\sqrt{3}}{2+\sqrt{3}}=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
4.
\(\frac{16-6\sqrt{7}}{\sqrt{7}-3}=\frac{3^2-2.3\sqrt{7}+7}{\sqrt{7}-3}=\frac{(\sqrt{7}-3)^2}{\sqrt{7}-3}=\sqrt{7}-3\)
5.
\(\frac{(\sqrt{3}-\sqrt{2})^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{2.3}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
6.
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{6-2\sqrt{10}}}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{6-2\sqrt{10}}}\)
\(=\sqrt{\dfrac{\left(5-2\sqrt{6}\right)^2}{25-24}+\sqrt{\left(3-\sqrt{6}\right)^2}}=\sqrt{25+24-20\sqrt{6}+3-\sqrt{6}}=\sqrt{52-21\sqrt{6}}\)