\(\sqrt{20} + \sqrt{80} = \sqrt{4} ⋅ \sqrt{5} + \sqrt{16} ⋅ \sqrt{5} = 2 \sqrt{5} + 4 \sqrt{5} \text{.}\)

\(2 \sqrt{5} + 4 \sqrt{5} = 6 \sqrt{5} \text{.}\)

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

\(-2 \sqrt{45} = -2 ⋅ \sqrt{9} ⋅ \sqrt{5} = -2 ⋅ 3 ⋅ \sqrt{5} = -6 \sqrt{5} \text{.}\)

\(6 \sqrt{32} - 3 \sqrt{8} = 6 ⋅ \sqrt{16} ⋅ \sqrt{2} - 3 ⋅ \sqrt{4} ⋅ \sqrt{2} \text{.}\)

\(6 ⋅ 4 ⋅ \sqrt{2} - 3 ⋅ 2 ⋅ \sqrt{2} = 24 \sqrt{2} - 6 \sqrt{2} = 18 \sqrt{2} \text{.}\)

\(\sqrt{1\frac{19}{81}} = \sqrt{\frac{100}{81}} = {\sqrt{100} \over \sqrt{81}} = \frac{10}{9} = 1\frac{1}{9} \text{.}\)

\({4 \over 3 \sqrt{2}} = {4 \over 3 \sqrt{2}} ⋅ {\sqrt{2} \over \sqrt{2}} = {4 \sqrt{2} \over 3 ⋅ 2} = \frac{2}{3} \sqrt{2} \text{.}\)

\(\sqrt{1\frac{37}{49}} = \sqrt{\frac{86}{49}} = {\sqrt{86} \over \sqrt{49}} = {\sqrt{86} \over 7} = \frac{1}{7} \sqrt{86} \text{.}\)

\(\sqrt{\frac{1}{13}} = {\sqrt{1} \over \sqrt{13}} = {1 \over \sqrt{13}} ⋅ {\sqrt{13} \over \sqrt{13}} = {\sqrt{13} \over 13} = \frac{1}{13} \sqrt{13} \text{.}\)

\(\sqrt{\frac{3}{32}} = {\sqrt{3} \over \sqrt{32}} ⋅ {\sqrt{32} \over \sqrt{32}} = {\sqrt{96} \over 32} = \frac{1}{32} \sqrt{96} = \frac{1}{32} ⋅ 4 ⋅ \sqrt{6} = \frac{1}{8} \sqrt{6} \text{.}\)

\({35 \sqrt{240} \over 7 \sqrt{6}} = {35 \over 7} ⋅ {\sqrt{240} \over \sqrt{6}} = 5 \sqrt{40} = 5 ⋅ \sqrt{4} ⋅ \sqrt{10} = 5 ⋅ 2 ⋅ \sqrt{10} = 10 \sqrt{10}\)

\(5 \sqrt{15} ⋅ 3 \sqrt{6} = 15 \sqrt{90} = 15 ⋅ \sqrt{9} ⋅ \sqrt{10} = 15 ⋅ 3 ⋅ \sqrt{10} = 45 \sqrt{10}\)