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28 tháng 6 2021
`(5sqrt7+7sqrt5):sqrt{35}`
`=(sqrt{5}.sqrt{7}.sqrt{5}+sqrt{5}.sqrt{7}.sqrt{7}):sqrt{35}`
`=sqrt{5}.sqrt{7}(sqrt5+sqrt7):sqrt{35}`
`=sqrt{35}(sqrt5+sqrt7):sqrt{35}`
`=sqrt5+sqrt7`
28 tháng 6 2021
Ta có : \(\left(\sqrt{5}\sqrt{5}\sqrt{7}+\sqrt{7}\sqrt{7}\sqrt{5}\right):\sqrt{35}\)
\(=\left(\sqrt{5}\sqrt{35}+\sqrt{7}\sqrt{35}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
23 tháng 8 2018
\(\dfrac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\dfrac{\sqrt{5}\left(\sqrt{5}.\sqrt{7}+7\right)}{\sqrt{35}}=\dfrac{\sqrt{7}\left(\sqrt{5}.\sqrt{7}+7\right)}{7}=\dfrac{7\sqrt{5}+7\sqrt{7}}{7}=\sqrt{5}+\sqrt{7}\)
24 tháng 7 2019
\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}=\left(\sqrt{5^2.7}+\sqrt{7^2.5}\right):\sqrt{35}\)
\(=\left(\sqrt{35.5}+\sqrt{35.7}\right):\sqrt{35}\)
\(=\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
Toán Học Team
1 tháng 7 2021
50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
AD
1
26 tháng 10 2023
\(\dfrac{\sqrt{35}-\sqrt{7}}{\sqrt{5}-1}-\dfrac{6}{\sqrt{7}-1}\)
\(=\dfrac{\sqrt{7}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{6\left(\sqrt{7}+1\right)}{6}\)
\(=\sqrt{7}-\sqrt{7}-1=-1\)
NH
\(\frac{\sqrt{20}-\sqrt{12}+2}{\sqrt{35}-\sqrt{21}+\sqrt{7}}+\frac{5}{\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
0
NQ
1
18 tháng 8 2020
1: Ta có: \(\sqrt{7-3\sqrt{5}}+\sqrt{7+3\sqrt{5}}\)
\(=\frac{\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{9-2\cdot3\cdot\sqrt{5}+5}+\sqrt{9+2\cdot3\cdot\sqrt{5}+5}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|3-\sqrt{5}\right|+\left|3+\sqrt{5}\right|}{\sqrt{2}}\)
\(=\frac{3-\sqrt{5}+3+\sqrt{5}}{\sqrt{2}}\)(Vì \(3>\sqrt{5}>0\))
\(=\frac{6}{\sqrt{2}}=\sqrt{18}=3\sqrt{2}\)
2) Ta có: \(\sqrt{6-\sqrt{35}}+\sqrt{6+\sqrt{35}}\)
\(=\frac{\sqrt{12-2\sqrt{35}}+\sqrt{12+2\sqrt{35}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{5}+5}+\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{5}+5}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-\sqrt{5}\right|+\left|\sqrt{7}+\sqrt{5}\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-\sqrt{5}+\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)(Vì \(\sqrt{7}>\sqrt{5}>0\))
\(=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
9 tháng 7 2019
\(\frac{\sqrt{7}+7}{\sqrt{7}+1}-\frac{\sqrt{7}-\sqrt{14}}{\sqrt{2}-1}+\frac{2\sqrt{35}-2\sqrt{7}}{1-\sqrt{5}}\)
\(=\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{\sqrt{7}+1}-\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{\sqrt{2}-1}+\frac{2\sqrt{7}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\)
\(=\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{\sqrt{7}+1}+\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\frac{2\sqrt{7}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\)
\(=\sqrt{7}+\sqrt{7}-2\sqrt{7}\)
\(=0\)
\(\dfrac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\dfrac{\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right)}{\sqrt{35}}=\sqrt{7}+\sqrt{5}\)