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Bài 2:
a: Ta có: \(\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{60}+6\right):2\sqrt{3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{12}\left(\sqrt{5}+\sqrt{3}\right):2\sqrt{3}\)
\(=2\sqrt{12}:2\sqrt{3}\)
=2
b: Ta có: \(\sqrt{5-\sqrt{21}}-\sqrt{\dfrac{7}{2}}\)
\(=\dfrac{\sqrt{10-2\sqrt{21}}-\sqrt{7}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}-\sqrt{7}}{\sqrt{2}}\)
\(=-\dfrac{\sqrt{6}}{2}\)
\(A=\dfrac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)+\sqrt{3}\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{5}+\sqrt{3}\)
ban xem lai de sai ko nhe
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)
`(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`
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
Lời giải:
\(=\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\frac{-\sqrt{5}(\sqrt{7}-\sqrt{3})}{\sqrt{7}-\sqrt{3}}=\frac{4(\sqrt{5}+1)}{5-1}-\sqrt{5}=(\sqrt{5}+1)-\sqrt{5}=1\)
\(\dfrac{4}{\sqrt{5}-1}+\dfrac{\sqrt{15}-\sqrt{35}}{\sqrt{7}-\sqrt{3}}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1
\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)
\(=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=-\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(=-\dfrac{\sqrt{21}}{7}\)
____________
\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{5}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{10}}{2}\)
Ta có: \(\dfrac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=1+\sqrt{3}+\sqrt{5}\)