\(\sqrt{3+4\sqrt{10+4\sqrt{6}}}=\sqrt{3+4\sqrt{\left(\sqrt{6}+2\right)^2}}=\sqrt{3+4\left(\sqrt{6}+2\right)}\)
\(=\sqrt{11+4\sqrt{6}}=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)
\(\sqrt{2-\sqrt{3}+\sqrt{2+\sqrt{3}}}\) hay là \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\) em nhỉ?
\(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\left(\sqrt{5}+1\right)}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
