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2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)
13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
=2
12.
\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)
13.
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|2\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)
`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`
`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`
`= sqrt3/sqrt 7`
`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`
`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`
`= sqrt5`
`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`
`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`
`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)
\(=\sqrt{6}+\sqrt{6}-1\)
\(=2\sqrt{6}-1\)
=======================
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)-3\left(\sqrt{2}-\sqrt{3}\right)}{-\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}\)
\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)
\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)
\(=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=-\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(=-\dfrac{\sqrt{21}}{7}\)
____________
\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{5}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{10}}{2}\)
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: \(=\sqrt{3}+1-\sqrt{3}=1\)
\(\dfrac{15}{\sqrt{6}-1}+\dfrac{8}{\sqrt{6}+2}+\dfrac{6}{3-\sqrt{6}}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{8\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\dfrac{6\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{6-1}+\dfrac{8\left(\sqrt{6}-2\right)}{6-4}+\dfrac{6\left(3+\sqrt{6}\right)}{9-6}-9\sqrt{6}\)
\(=3\left(\sqrt{6}+1\right)+4\left(\sqrt{6}-2\right)+2\left(3+\sqrt{6}\right)-9\sqrt{6}\)
\(=3\sqrt{6}+3+4\sqrt{6}-8+6+2\sqrt{6}-9\sqrt{6}\)
\(=9\sqrt{6}+1-9\sqrt{6}\)
\(=1\)
\(\sqrt{\left(\sqrt{5}-1\right)\sqrt{14-6\sqrt{5}}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{9-6\sqrt{5}+5}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left|3-\sqrt{5}\right|}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{3\sqrt{5}-5-3+\sqrt{5}}\)
\(=\sqrt{4\sqrt{5}-8}\)
\(=\sqrt{4\left(\sqrt{5}-2\right)}\)
\(=2\sqrt{\sqrt{5}-2}\)
`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`
`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`
`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`
`=-(7-5)=-2`
`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`
`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`
`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`
`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`
`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`
`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`
a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=-2\)
b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)
\(=\sqrt{5}\)
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
\(\dfrac{\sqrt{\sqrt{15}-\sqrt{6}}+\sqrt{\sqrt{15}+\sqrt{6}}}{\sqrt{\sqrt{15}+3}}\)
\(=\dfrac{\sqrt{\sqrt{3}.\left(\sqrt{5}-\sqrt{3}\right)}+\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}{\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\sqrt{3}}.\left[\left(\sqrt{5}-\sqrt{3}\right)+\left(\sqrt{5}+\sqrt{3}\right)\right]}{\sqrt{\sqrt{3}.\left(\sqrt{5}+\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{2\sqrt{5}}{\sqrt{5}+\sqrt{3}}\)
Chúc bạn học tốt!!!