\(\sqrt{20}=\sqrt{4}⋅\sqrt{5}=2\sqrt{5}\text{.}\)

\(3\sqrt{300}=3⋅\sqrt{100}⋅\sqrt{3}=3⋅10⋅\sqrt{3}=30\sqrt{3}\text{.}\)

\(\sqrt{\frac{1}{81}}={\sqrt{1} \over \sqrt{81}}=\frac{1}{9}\text{.}\)

\(\sqrt{1\frac{23}{25}}=\sqrt{\frac{48}{25}}={\sqrt{48} \over \sqrt{25}}={\sqrt{48} \over 5}=\frac{1}{5}\sqrt{48}=\frac{1}{5}⋅4⋅\sqrt{3}=\frac{4}{5}\sqrt{3}\text{.}\)

\(\sqrt{32}+\sqrt{50}=\sqrt{16}⋅\sqrt{2}+\sqrt{25}⋅\sqrt{2}=4\sqrt{2}+5\sqrt{2}\text{.}\)

\(4\sqrt{2}+5\sqrt{2}=9\sqrt{2}\text{.}\)

\(3\sqrt{12}-5\sqrt{48}=3⋅\sqrt{4}⋅\sqrt{3}-5⋅\sqrt{16}⋅\sqrt{3}\text{.}\)

\(3⋅2⋅\sqrt{3}-5⋅4⋅\sqrt{3}=6\sqrt{3}-20\sqrt{3}=-14\sqrt{3}\text{.}\)

\({6 \over 5\sqrt{7}}={6 \over 5\sqrt{7}}⋅{\sqrt{7} \over \sqrt{7}}={6\sqrt{7} \over 5⋅7}=\frac{6}{35}\sqrt{7}\text{.}\)

\(\sqrt{1\frac{29}{35}}=\sqrt{\frac{64}{35}}={\sqrt{64} \over \sqrt{35}}={8 \over \sqrt{35}}⋅{\sqrt{35} \over \sqrt{35}}={8\sqrt{35} \over 35}=\frac{8}{35}\sqrt{35}\text{.}\)

\(\sqrt{\frac{7}{50}}={\sqrt{7} \over \sqrt{50}}⋅{\sqrt{50} \over \sqrt{50}}={\sqrt{350} \over 50}=\frac{1}{50}\sqrt{350}=\frac{1}{50}⋅5⋅\sqrt{14}=\frac{1}{10}\sqrt{14}\text{.}\)