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\(B=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\right)\)
\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
\(=\dfrac{1}{3\sqrt{2}}\cdot\dfrac{6}{1}=\sqrt{2}\)
\(\sqrt{3}.\sqrt{\frac{1}{27}}.\sqrt{5}.\sqrt{20}=\sqrt{3.\frac{1}{27}}.\sqrt{5.20}=\sqrt{\frac{1}{9}}.\sqrt{100}=\frac{1}{3}.10=\frac{10}{3}\)
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)