\(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
K
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TK
1
11 tháng 9 2023
\(=\dfrac{2\cdot\sqrt{3+\sqrt{5}-2\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\cdot\sqrt{2-2\sqrt{3}+\sqrt{5}}}{\sqrt{6}+\sqrt{2}}\)
PD
1
11 tháng 12 2017
\(A=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{8-\sqrt{48}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6-2\sqrt{6.2}+2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=1\)
TH
1
D
20 tháng 7 2017
Ta có C=\(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
=\(\frac{2\sqrt{3-\sqrt{3+\sqrt{1+2\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
=\(\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)}^2}}}{\sqrt{6}-\sqrt{2}}\)
=\(\frac{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)=\(\frac{2\sqrt{3-\sqrt{\left(1+\sqrt{3}\right)}^2}}{\sqrt{6}-\sqrt{2}}\)=\(\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=1
28 tháng 7 2020
Ta có: \(B=\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+3\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+3\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+3\cdot\left(\sqrt{2}+1\right)}\)
\(=\sqrt{13+3\sqrt{2}+3}\)
\(=\sqrt{16+3\sqrt{2}}\)
Ta có: \(C=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\left(\sqrt{3}+1\right)^2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{\sqrt{3}-1}=1\)
Ta có: \(D=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-2}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{1}}\)
\(=\sqrt{6+2\sqrt{2}}\)
L
0
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{2}.\sqrt{4-2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=1\)