Tinh:
\(\sqrt{2-\sqrt{3}}\)
K
Khách
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3N
0
13 tháng 12 2022
\(B=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\right)\)
\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
\(=\dfrac{1}{3\sqrt{2}}\cdot\dfrac{6}{1}=\sqrt{2}\)
ML
25 tháng 6 2015
Ta có: \(4\left(1+\frac{\sqrt{3}}{2}\right)=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\Rightarrow1+\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}+1}{2}\right)^2\)
Tương tự \(1-\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}-1}{2}\right)^2\)
\(VT=\frac{\left(\frac{\sqrt{3}+1}{2}\right)^2}{1+\frac{\sqrt{3}+1}{2}}+\frac{\left(\frac{\sqrt{3}-1}{2}\right)^2}{1-\frac{\sqrt{3}-1}{2}}=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{3+\sqrt{3}}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{3-\sqrt{3}}{2}}\)\(=\frac{\left(\sqrt{3}+1\right)^2}{2.\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2.\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{2\sqrt{3}}+\frac{\sqrt{3}-1}{2\sqrt{3}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2\sqrt{3}}=1=VP\)
L
Tinh \(\sqrt{5}\left(\sqrt{6}+1\right):\frac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}-\sqrt{2}}}\)
3
PL
0
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
KQ
1
25 tháng 8 2019
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}.1}+1}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\left|\sqrt{3}+1\right|}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
=\(\sqrt{6+2.\left(\sqrt{2}.\sqrt{2-\sqrt{3}}\right)}\)
=\(\sqrt{6+2.\left(\sqrt{4-2\sqrt{3}}\right)}\)
=\(\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6+2.\left|\sqrt{3}-1\right|}\)
=\(\sqrt{6+2\sqrt{3}-2}\)
=\(\sqrt{4+2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)
=\(\left|\sqrt{3}+1\right|\)
=\(\sqrt{3}+1\)
=0.5176380902
study well
k nha'
ai k đúng cho mk mk trả lại gấp đôi