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= \(\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2+\sqrt{2}\right)^3}\) = \(2+\sqrt{2}+2+\sqrt{2}\) = 4+\(2\sqrt{2}\)
b)\(27-10\sqrt{2}=5^2-2.5\sqrt{2}+2=\left(5-\sqrt{2}\right)^2\)
c)\(18-8\sqrt{2}=4^2-2.4\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)
d)\(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
e)\(6\sqrt{5}+14=9+2.3\sqrt{5}+5=\left(3+\sqrt{5}\right)^2\)
f)\(20\sqrt{5}+45=5^2+2.5.2\sqrt{5}+20=\left(5+2\sqrt{5}\right)^2\)
g)\(7-2\sqrt{6}=6-2\sqrt{6}+1=\left(\sqrt{6}-1\right)^2\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
= 5
\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)
\(=3\)
\(A=\sqrt[3]{\left(\frac{1}{2}+\frac{1}{2}\sqrt{13}\right)^3}+\sqrt[3]{\left(\frac{1}{2}-\frac{1}{2}\sqrt{13}\right)^3}\)
\(=\frac{1}{2}+\frac{\sqrt{13}}{2}+\frac{1}{2}-\frac{\sqrt{13}}{2}=1\)
\(B=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4\)