\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)

\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)

Mình sửa lại chút nha bạn

\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)

\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)

\(=\dfrac{26\sqrt{5}+8\sqrt{2}+36+25+11\sqrt{3}}{6}\)

\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+8\sqrt{2}}{6}\)

a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)

\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)

\(=\sqrt{518400}\)

=720

b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)

\(=9-2\sqrt{14}+2\sqrt{14}\)

=9

c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)

\(=6+2\sqrt{5}-6\)

\(=2\sqrt{5}\)

6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

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Rất muốn tính nhưng mà dài quá

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)

\(\sqrt{5}+2\sqrt{6}-\left(\sqrt{5}-2\sqrt{6}\right)=4\sqrt{6}\)

`\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}`

`=\sqrt{2+2.\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2.\sqrt{2}.\sqrt{3}+3}`

`=\sqrt{(\sqrt{2}+\sqrt{3})^{2}}-\sqrt{(\sqrt{2}-\sqrt{3})^{2}}`

`=|\sqrt{2}+\sqrt{3}|-|\sqrt{2}-\sqrt{3}|`

`=\sqrt{2}+\sqrt{3}+\sqrt{2}-\sqrt{3}`

`=2\sqrt{2}`

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)