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

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

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

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

\(5 \sqrt{8} - 4 \sqrt{18} = 5 ⋅ \sqrt{4} ⋅ \sqrt{2} - 4 ⋅ \sqrt{9} ⋅ \sqrt{2} \text{.}\)

\(5 ⋅ 2 ⋅ \sqrt{2} - 4 ⋅ 3 ⋅ \sqrt{2} = 10 \sqrt{2} - 12 \sqrt{2} = -2 \sqrt{2} \text{.}\)

\(\sqrt{1\frac{15}{49}} = \sqrt{\frac{64}{49}} = {\sqrt{64} \over \sqrt{49}} = \frac{8}{7} = 1\frac{1}{7} \text{.}\)

\({9 \over 7 \sqrt{2}} = {9 \over 7 \sqrt{2}} ⋅ {\sqrt{2} \over \sqrt{2}} = {9 \sqrt{2} \over 7 ⋅ 2} = \frac{9}{14} \sqrt{2} \text{.}\)

\(\sqrt{10\frac{2}{9}} = \sqrt{\frac{92}{9}} = {\sqrt{92} \over \sqrt{9}} = {\sqrt{92} \over 3} = \frac{1}{3} \sqrt{92} = \frac{1}{3} ⋅ 2 ⋅ \sqrt{23} = \frac{2}{3} \sqrt{23} \text{.}\)

\(\sqrt{\frac{16}{85}} = {\sqrt{16} \over \sqrt{85}} = {4 \over \sqrt{85}} ⋅ {\sqrt{85} \over \sqrt{85}} = {4 \sqrt{85} \over 85} = \frac{4}{85} \sqrt{85} \text{.}\)

\(\sqrt{\frac{7}{11}} = {\sqrt{7} \over \sqrt{11}} ⋅ {\sqrt{11} \over \sqrt{11}} = {\sqrt{77} \over 11} = \frac{1}{11} \sqrt{77} \text{.}\)

\({36 \sqrt{100} \over 9 \sqrt{5}} = {36 \over 9} ⋅ {\sqrt{100} \over \sqrt{5}} = 4 \sqrt{20} = 4 ⋅ \sqrt{4} ⋅ \sqrt{5} = 4 ⋅ 2 ⋅ \sqrt{5} = 8 \sqrt{5}\)

\(3 \sqrt{2} ⋅ 2 \sqrt{10} = 6 \sqrt{20} = 6 ⋅ \sqrt{4} ⋅ \sqrt{5} = 6 ⋅ 2 ⋅ \sqrt{5} = 12 \sqrt{5}\)