3.
a. $2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}$
$=2\sqrt{6}-4\sqrt{2}+(1+8+4\sqrt{2})-2\sqrt{6}$
$=2\sqrt{6}-2\sqrt{6}-4\sqrt{2}+4\sqrt{2}+9$
$=9$
b.
$\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}$
$=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}$
$=\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}$
$=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}$
c.
\(\sqrt{\frac{4}{(2-\sqrt{5})^2}}-\sqrt{\frac{4}{(2+\sqrt{5})^2}}=\sqrt{(\frac{2}{2-\sqrt{5}})^2}-\sqrt{(\frac{2}{(2+\sqrt{5}})^2}\)
\(=|\frac{2}{2-\sqrt{5}}|-|\frac{2}{2+\sqrt{5}}|=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2(\sqrt{5}+2)-2(\sqrt{5}-2)}{(\sqrt{5}-2)(\sqrt{5}+2)}=\frac{8}{5-2^2}=8\)
Bài 3:
a) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(2\sqrt{2}+1\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}-2\sqrt{6}+9+4\sqrt{2}\)
=9
b) Ta có: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\sqrt{6}\)
Bài 2:
a) Ta có: \(A=\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
