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

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

\(\sqrt{200} = \sqrt{100} ⋅ \sqrt{2} = 10 \sqrt{2} \text{.}\)

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

\(7 \sqrt{80} - 2 \sqrt{125} = 7 ⋅ \sqrt{16} ⋅ \sqrt{5} - 2 ⋅ \sqrt{25} ⋅ \sqrt{5} \text{.}\)

\(7 ⋅ 4 ⋅ \sqrt{5} - 2 ⋅ 5 ⋅ \sqrt{5} = 28 \sqrt{5} - 10 \sqrt{5} = 18 \sqrt{5} \text{.}\)

\(\sqrt{\frac{64}{81}} = {\sqrt{64} \over \sqrt{81}} = \frac{8}{9} \text{.}\)

\({7 \over 8 \sqrt{3}} = {7 \over 8 \sqrt{3}} ⋅ {\sqrt{3} \over \sqrt{3}} = {7 \sqrt{3} \over 8 ⋅ 3} = \frac{7}{24} \sqrt{3} \text{.}\)

\(\sqrt{5\frac{2}{9}} = \sqrt{\frac{47}{9}} = {\sqrt{47} \over \sqrt{9}} = {\sqrt{47} \over 3} = \frac{1}{3} \sqrt{47} \text{.}\)

\(\sqrt{1\frac{1}{99}} = \sqrt{\frac{100}{99}} = {\sqrt{100} \over \sqrt{99}} = {10 \over \sqrt{99}} ⋅ {\sqrt{99} \over \sqrt{99}} = {10 \sqrt{99} \over 99} = \frac{10}{99} \sqrt{99} = \frac{10}{99} ⋅ 3 ⋅ \sqrt{11} = \frac{10}{33} \sqrt{11} \text{.}\)

\(\sqrt{\frac{5}{12}} = {\sqrt{5} \over \sqrt{12}} ⋅ {\sqrt{12} \over \sqrt{12}} = {\sqrt{60} \over 12} = \frac{1}{12} \sqrt{60} = \frac{1}{12} ⋅ 2 ⋅ \sqrt{15} = \frac{1}{6} \sqrt{15} \text{.}\)

\({27 \sqrt{240} \over 9 \sqrt{6}} = {27 \over 9} ⋅ {\sqrt{240} \over \sqrt{6}} = 3 \sqrt{40} = 3 ⋅ \sqrt{4} ⋅ \sqrt{10} = 3 ⋅ 2 ⋅ \sqrt{10} = 6 \sqrt{10}\)

\(3 \sqrt{3} ⋅ 2 \sqrt{15} = 6 \sqrt{45} = 6 ⋅ \sqrt{9} ⋅ \sqrt{5} = 6 ⋅ 3 ⋅ \sqrt{5} = 18 \sqrt{5}\)