a) Ta có: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\cdot\left(2-\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-\left(\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-\left|2+\sqrt{3}\right|}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)\cdot\left(2-\sqrt{3}\right)}\)
\(=\sqrt{2^2-\left(\sqrt{3}\right)^2}\)
\(=\sqrt{4-3}=\sqrt{1}=1\)
b)Sửa đề: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{\sqrt{3}+2}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{\sqrt{3}+2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\cdot\sqrt{2\sqrt{3}+4}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}+1\right)^2\cdot\left(\sqrt{3}-2\right)\)
\(=\left(3+2\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\)
\(=3\sqrt{3}-6+6-4\sqrt{3}+\sqrt{3}-2\)
\(=-2\)
c) Ta có: \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(=\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
\(=\frac{\sqrt{42-2\cdot\sqrt{42}\cdot\sqrt{6}+6}}{2}-\frac{\sqrt{42+2\cdot\sqrt{42}\cdot\sqrt{6}+6}}{2}\)
\(=\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)}^2}{2}\)
\(=\frac{\left|\sqrt{42}-\sqrt{6}\right|-\left|\sqrt{42}+\sqrt{6}\right|}{2}\)
\(=\frac{\left(\sqrt{42}-\sqrt{6}-\sqrt{42}-\sqrt{6}\right)}{2}=\frac{-2\sqrt{6}}{2}=-\sqrt{6}\)
