Rút gọn:
\(A=\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
\(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\)
\(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)
\(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(A=\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
\(A=\dfrac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+2\sqrt{2}\right)}+\dfrac{\sqrt{2}.\left(\sqrt{3}+3\right)}{\sqrt{2}.\left(\sqrt{2+\sqrt{3}}-2\sqrt{2}\right)}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{4-2\sqrt{3}}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{4+2\sqrt{3}}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{3}-1+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}+1-4}\)
\(A=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{6}\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}+\dfrac{\sqrt{3}\left(\sqrt{2}+\sqrt{6}\right)}{\sqrt{3}\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}}{1+\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{6}}{1-\sqrt{3}}=\dfrac{\left(\sqrt{2}-\sqrt{6}\right)\left(1-\sqrt{3}\right)+\left(\sqrt{2}+\sqrt{6}\right)\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}-\sqrt{6}+3\sqrt{2}+\sqrt{2}+\sqrt{6}+\sqrt{6}+3\sqrt{2}}{1-3}=\dfrac{8\sqrt{2}}{-2}=-4\sqrt{2}\)
* \(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\) \(=\dfrac{\sqrt{6+2.\sqrt{6}.\sqrt{5}+5}-\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}{\sqrt{5}}\)\(=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}}{\sqrt{5}}=\dfrac{2\sqrt{5}}{\sqrt{5}}=2\)
* \(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)
Đặt:\(y=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\Rightarrow y^2=4+\sqrt{15}+4-\sqrt{15}+2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=8+2=10\Rightarrow y=\sqrt{10}\)
Suy ra: \(C=\sqrt{12+4\sqrt{5}}-y=\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}-\sqrt{10}=\sqrt{10}+\sqrt{2}-\sqrt{10}=\sqrt{2}\)* \(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)+\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{1}=4\)
