\(\sqrt{125}=\sqrt{25}⋅\sqrt{5}=5\sqrt{5}\text{.}\)

\(-7\sqrt{12}=-7⋅\sqrt{4}⋅\sqrt{3}=-7⋅2⋅\sqrt{3}=-14\sqrt{3}\text{.}\)

\(\sqrt{\frac{25}{64}}={\sqrt{25} \over \sqrt{64}}=\frac{5}{8}\text{.}\)

\(\sqrt{2\frac{1}{49}}=\sqrt{\frac{99}{49}}={\sqrt{99} \over \sqrt{49}}={\sqrt{99} \over 7}=\frac{1}{7}\sqrt{99}=\frac{1}{7}⋅3⋅\sqrt{11}=\frac{3}{7}\sqrt{11}\text{.}\)

\(\sqrt{45}+\sqrt{20}=\sqrt{9}⋅\sqrt{5}+\sqrt{4}⋅\sqrt{5}=3\sqrt{5}+2\sqrt{5}\text{.}\)

\(3\sqrt{5}+2\sqrt{5}=5\sqrt{5}\text{.}\)

\(2\sqrt{45}+6\sqrt{125}=2⋅\sqrt{9}⋅\sqrt{5}+6⋅\sqrt{25}⋅\sqrt{5}\text{.}\)

\(2⋅3⋅\sqrt{5}+6⋅5⋅\sqrt{5}=6\sqrt{5}+30\sqrt{5}=36\sqrt{5}\text{.}\)

\({3 \over 2\sqrt{3}}={3 \over 2\sqrt{3}}⋅{\sqrt{3} \over \sqrt{3}}={3\sqrt{3} \over 2⋅3}=\frac{1}{2}\sqrt{3}\text{.}\)

\(\sqrt{\frac{9}{40}}={\sqrt{9} \over \sqrt{40}}={3 \over \sqrt{40}}⋅{\sqrt{40} \over \sqrt{40}}={3\sqrt{40} \over 40}=\frac{3}{40}\sqrt{40}=\frac{3}{40}⋅2⋅\sqrt{10}=\frac{3}{20}\sqrt{10}\text{.}\)

\(\sqrt{\frac{2}{33}}={\sqrt{2} \over \sqrt{33}}⋅{\sqrt{33} \over \sqrt{33}}={\sqrt{66} \over 33}=\frac{1}{33}\sqrt{66}\text{.}\)