Tính \(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
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TT
4
7 tháng 6 2017
Giải từ từ lần lượt các biểu thức trong dấu căn nhé:
\(\sqrt{13+\sqrt{48}}=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}=\sqrt{\left(2\sqrt{3}+1\right)^2}=2\sqrt{3}+1\)
\(\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(B=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{3+2\sqrt{3}+1}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\)
7 tháng 6 2017
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+12}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-1-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{1-2\sqrt{3}+\sqrt{3}^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(1-\sqrt{3}\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{6-2}\)
\(\frac{\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
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) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
TK
1
11 tháng 9 2023
\(=\dfrac{2\cdot\sqrt{3+\sqrt{5}-2\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\cdot\sqrt{2-2\sqrt{3}+\sqrt{5}}}{\sqrt{6}+\sqrt{2}}\)
3 tháng 2 2022
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)