+) ta có \(A=\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(\dfrac{5+2\sqrt{6}+10-4\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(\dfrac{\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\dfrac{15^2-\left(2\sqrt{6}\right)^2}{5^2-\left(2\sqrt{6}\right)^2}=201\)
+) \(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}\sqrt{\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt{2+\sqrt{2}}\right)}\)
\(=\sqrt{4+\sqrt{8}}\sqrt{2^2-\left(\sqrt{2+\sqrt{2}}\right)^2}=\sqrt{2}\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2}\left(\sqrt{2^2-\left(\sqrt{2}\right)^2}\right)=\sqrt{2}.\sqrt{2}=2\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt{2+\sqrt{2}}\right)}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{4-2-\sqrt{2}}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2}}\)
\(B=\sqrt{8-4\sqrt{2}+4\sqrt{2}-4}=\sqrt{4}=2\)
