a, đặt \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(b,\)
\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1=2
b) Ta có: \(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=4\sqrt{7}\)