Ta có: \(\sqrt{6-3\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{3-\sqrt{3}+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(=\sqrt{3\left(2-\sqrt{3}\right)}+\sqrt{2-\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{3}+1\right)}{\sqrt{2}}\sqrt{4-2\sqrt{3}}=\dfrac{\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(\sqrt{6-3\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left[\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right]\)
\(=\dfrac{1}{\sqrt{2}}\left(3-\sqrt{3}+\sqrt{3}-1\right)\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
