\(\sqrt{12} + \sqrt{48} = \sqrt{4} ⋅ \sqrt{3} + \sqrt{16} ⋅ \sqrt{3} = 2 \sqrt{3} + 4 \sqrt{3} \text{.}\)

\(2 \sqrt{3} + 4 \sqrt{3} = 6 \sqrt{3} \text{.}\)

\(\sqrt{8} = \sqrt{4} ⋅ \sqrt{2} = 2 \sqrt{2} \text{.}\)

\(7 \sqrt{8} = 7 ⋅ \sqrt{4} ⋅ \sqrt{2} = 7 ⋅ 2 ⋅ \sqrt{2} = 14 \sqrt{2} \text{.}\)

\(5 \sqrt{48} - 7 \sqrt{27} = 5 ⋅ \sqrt{16} ⋅ \sqrt{3} - 7 ⋅ \sqrt{9} ⋅ \sqrt{3} \text{.}\)

\(5 ⋅ 4 ⋅ \sqrt{3} - 7 ⋅ 3 ⋅ \sqrt{3} = 20 \sqrt{3} - 21 \sqrt{3} = -1 \sqrt{3} \text{.}\)

\(\sqrt{\frac{16}{81}} = {\sqrt{16} \over \sqrt{81}} = \frac{4}{9} \text{.}\)

\({9 \over 4 \sqrt{2}} = {9 \over 4 \sqrt{2}} ⋅ {\sqrt{2} \over \sqrt{2}} = {9 \sqrt{2} \over 4 ⋅ 2} = 1\frac{1}{8} \sqrt{2} \text{.}\)

\(\sqrt{\frac{12}{25}} = {\sqrt{12} \over \sqrt{25}} = {\sqrt{12} \over 5} = \frac{1}{5} \sqrt{12} = \frac{1}{5} ⋅ 2 ⋅ \sqrt{3} = \frac{2}{5} \sqrt{3} \text{.}\)

\(\sqrt{4\frac{1}{6}} = \sqrt{\frac{25}{6}} = {\sqrt{25} \over \sqrt{6}} = {5 \over \sqrt{6}} ⋅ {\sqrt{6} \over \sqrt{6}} = {5 \sqrt{6} \over 6} = \frac{5}{6} \sqrt{6} \text{.}\)

\(\sqrt{22\frac{1}{2}} = \sqrt{\frac{45}{2}} = {\sqrt{45} \over \sqrt{2}} ⋅ {\sqrt{2} \over \sqrt{2}} = {\sqrt{90} \over 2} = \frac{1}{2} \sqrt{90} = \frac{1}{2} ⋅ 3 ⋅ \sqrt{10} = 1\frac{1}{2} \sqrt{10} \text{.}\)

\({12 \sqrt{240} \over 2 \sqrt{10}} = {12 \over 2} ⋅ {\sqrt{240} \over \sqrt{10}} = 6 \sqrt{24} = 6 ⋅ \sqrt{4} ⋅ \sqrt{6} = 6 ⋅ 2 ⋅ \sqrt{6} = 12 \sqrt{6}\)

\(3 \sqrt{10} ⋅ 4 \sqrt{6} = 12 \sqrt{60} = 12 ⋅ \sqrt{4} ⋅ \sqrt{15} = 12 ⋅ 2 ⋅ \sqrt{15} = 24 \sqrt{15}\)