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 \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

rộp rộp

a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)

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

\(49+20\sqrt{6}=25+2.5.2\sqrt{6}+24=\left(5+2\sqrt{6}\right)^2=\left(3+2.\sqrt{3}\sqrt{2}+2\right)^2=\left(\sqrt{3}+\sqrt{2}\right)^4\)

\(\Leftrightarrow\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

tuiwng tự \(\Leftrightarrow\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\)

=> Cộng lại  = > dpcm

\(\dfrac{1}{\sqrt{49+20\sqrt{6}}}-\dfrac{1}{\sqrt{49-20\sqrt{6}}}+\dfrac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{5^2+2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{5^2-2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)

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

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

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

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

\(=-4\sqrt{6}+2+\sqrt{3}\)

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

\(=5-2\sqrt{6}-5-2\sqrt{6}+2+\sqrt{3}\)

\(=2-4\sqrt{6}+\sqrt{3}\)

\(=\dfrac{1}{\sqrt{\sqrt{25}^2+2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}-\dfrac{1}{\sqrt{\sqrt{25}^2-2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}+\dfrac{1}{\sqrt{\sqrt{4}^2-2\sqrt{4}\sqrt{3}+\sqrt{3}^2}}\\ =\dfrac{1}{\sqrt{\left(\sqrt{25}+\sqrt{24}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{25}-\sqrt{24}\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{4}-\sqrt{3}\right)^2}}\\ =\dfrac{1}{5+\sqrt{24}}-\dfrac{1}{5-\sqrt{24}}+\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{5-\sqrt{24}}{25-24}-\dfrac{5+\sqrt{24}}{25-24}+\dfrac{2+\sqrt{3}}{4-3}\\ =5-\sqrt{24}-5-\sqrt{24}+2+\sqrt{3}\\ =2-4\sqrt{6}+\sqrt{3}\)

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

Trả lời:

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

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

\(A=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

\(A=\sqrt{1}\)

\(A=1\)

\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=1\)

a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)

\(=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)

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

\(=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)( đpcm )

\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)

\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)

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