CMR : \(\sqrt[4]{49+20\times\sqrt{6}}+\sqrt[4]{49-20\times\sqrt{6}}=2\times\sqrt{3}\)
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5 tháng 5 2016
Ta có \(\sqrt[4]{49+20\sqrt{6}}=\sqrt[4]{25+10\sqrt{24}+24}=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}\)
\(=\sqrt[4]{\left(\sqrt{3}+\sqrt{2}\right)^4}=\sqrt{3}+\sqrt{2}\)
Tương tự : \(\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\) ( Do \(\sqrt{3}>\sqrt{2}\) )
Suy ra \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)
NC
3
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
\(\dfrac{1}{\sqrt{49+20\sqrt{6}}}-\dfrac{1}{\sqrt{49-20\sqrt{6}}}+\dfrac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\dfrac{1}{\sqrt{5^2+2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{5^2-2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{1}{\sqrt{\left(5+2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{\left(5-2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{\left(2-\sqrt{3}\right)^2}}\)
\(=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}+\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{5-2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}-\dfrac{5+2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{1}+\dfrac{2+\sqrt{3}}{1}\)
\(=-4\sqrt{6}+2+\sqrt{3}\)
24 tháng 8 2023
\(=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}+\dfrac{1}{2-\sqrt{3}}\)
\(=5-2\sqrt{6}-5-2\sqrt{6}+2+\sqrt{3}\)
\(=2-4\sqrt{6}+\sqrt{3}\)
RR
2
25 tháng 7 2020
Trả lời:
\(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)
Ta có:\(VT=\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}\)
\(=\sqrt[4]{25+20\sqrt{6}+24}+\sqrt[4]{25-20\sqrt{6}+24}\)
\(=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}+\sqrt[4]{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)
\(=2\sqrt{3}=VP\)
Vậy \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)
CC
8 tháng 10 2017
a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)
KQ
1
17 tháng 7 2019
49 + 20 căn 6 = 25 + 2.5.(2 căn 6) +24 = (5 + 2 căn 6)2
tương tự vs 49 - 20 căn 6 = (5 - 2 căn 6)2 =) căn ( 49 - 20 căn 6 ) = 5 - 2 căn 6
7 - 4 căn 3 = 4 - 4 căn 3 + 3 = (2 - căn 3)2 =) căn ( 7 - 4 căn 3 ) = 2 - căn 3
tự giải nhé
HH
15 tháng 6 2018
Giải:
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)
\(=\sqrt{2}\)
Vậy ...
5 tháng 8 2021
\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
\(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)= \(2\sqrt{3}\)
\(49+20\sqrt{6}=25+2.5.2\sqrt{6}+24=\left(5+2\sqrt{6}\right)^2=\left(3+2.\sqrt{3}\sqrt{2}+2\right)^2=\left(\sqrt{3}+\sqrt{2}\right)^4\)
\(\Leftrightarrow\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
tuiwng tự \(\Leftrightarrow\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
=> Cộng lại = > dpcm