\(A=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
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TK
1
11 tháng 9 2023
\(=\dfrac{2\cdot\sqrt{3+\sqrt{5}-2\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\cdot\sqrt{2-2\sqrt{3}+\sqrt{5}}}{\sqrt{6}+\sqrt{2}}\)
V
1
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 10 2019
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{2}.\sqrt{4-2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=1\)
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
Q
25 tháng 6 2017
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)
\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
DT
1
3 tháng 8 2017
Ta có :
A= \(\dfrac{2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Đặt B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Ta có B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}}\)
\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12}-1}}\\ =2\sqrt{3+\sqrt{4-\sqrt{12}}}\\ =2\cdot\sqrt{3+\sqrt{3-2\cdot\sqrt{3}+1}}\\ =2\cdot\sqrt{3+\sqrt{3}-1}\\ =2\cdot\sqrt{2+\sqrt{3}}\)
Thay B vào A, ta cũng có:
A=\(\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\ =\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2\cdot\left(\sqrt{3}+1\right)}}\\ =\dfrac{\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}=1\)
Vậy A thuộc Z
\(A=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{8-\sqrt{48}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6-2\sqrt{6.2}+2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=1\)