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\(\sqrt{1\frac{11}{25}} = \sqrt{\frac{36}{25}} = {\sqrt{36} \over \sqrt{25}} = \frac{6}{5} = 1\frac{1}{5} \text{.}\)

\(\sqrt{3\frac{21}{25}} = \sqrt{\frac{96}{25}} = {\sqrt{96} \over \sqrt{25}} = {\sqrt{96} \over 5} = \frac{1}{5} \sqrt{96} = \frac{1}{5} ⋅ 4 ⋅ \sqrt{6} = \frac{4}{5} \sqrt{6} \text{.}\)

\(\sqrt{\frac{81}{92}} = {\sqrt{81} \over \sqrt{92}} = {9 \over \sqrt{92}} ⋅ {\sqrt{92} \over \sqrt{92}} = {9 \sqrt{92} \over 92} = \frac{9}{92} \sqrt{92} = \frac{9}{92} ⋅ 2 ⋅ \sqrt{23} = \frac{9}{46} \sqrt{23} \text{.}\)

\(\sqrt{10\frac{2}{3}} = \sqrt{\frac{32}{3}} = {\sqrt{32} \over \sqrt{3}} ⋅ {\sqrt{3} \over \sqrt{3}} = {\sqrt{96} \over 3} = \frac{1}{3} \sqrt{96} = \frac{1}{3} ⋅ 4 ⋅ \sqrt{6} = 1\frac{1}{3} \sqrt{6} \text{.}\)

\({28 \sqrt{56} \over 4 \sqrt{7}} = {28 \over 4} ⋅ {\sqrt{56} \over \sqrt{7}} = 7 \sqrt{8} = 7 ⋅ \sqrt{4} ⋅ \sqrt{2} = 7 ⋅ 2 ⋅ \sqrt{2} = 14 \sqrt{2}\)

\(\sqrt{112} = \sqrt{16} ⋅ \sqrt{7} = 4 \sqrt{7} \text{.}\)

\(2 \sqrt{63} = 2 ⋅ \sqrt{9} ⋅ \sqrt{7} = 2 ⋅ 3 ⋅ \sqrt{7} = 6 \sqrt{7} \text{.}\)

\(\sqrt{18} + \sqrt{50} = \sqrt{9} ⋅ \sqrt{2} + \sqrt{25} ⋅ \sqrt{2} = 3 \sqrt{2} + 5 \sqrt{2} \text{.}\)

\(3 \sqrt{2} + 5 \sqrt{2} = 8 \sqrt{2} \text{.}\)

\(6 \sqrt{18} - 3 \sqrt{32} = 6 ⋅ \sqrt{9} ⋅ \sqrt{2} - 3 ⋅ \sqrt{16} ⋅ \sqrt{2} \text{.}\)

\(6 ⋅ 3 ⋅ \sqrt{2} - 3 ⋅ 4 ⋅ \sqrt{2} = 18 \sqrt{2} - 12 \sqrt{2} = 6 \sqrt{2} \text{.}\)

\({6 \over 4 + \sqrt{3}} = {6 \over 4 + \sqrt{3}} ⋅ {4 - \sqrt{3} \over 4 - \sqrt{3}}\)
\(\text{} = {6 (4 + \sqrt{3}) \over 16 - 3}\)
\(\text{} = \frac{6}{13} (4 + \sqrt{3})\)
\(\text{} = 1\frac{11}{13} + \frac{6}{13} \sqrt{3}\)

\({\sqrt{2} \over \sqrt{3} - \sqrt{5}} = {\sqrt{2} \over \sqrt{3} - \sqrt{5}} ⋅ {\sqrt{3} + \sqrt{5} \over \sqrt{3} + \sqrt{5}}\)
\(\text{} = {\sqrt{2} (\sqrt{3} + \sqrt{5}) \over 3 - 5}\)
\(\text{} = -\frac{1}{2} \sqrt{2} (\sqrt{3} + \sqrt{5})\)
\(\text{} = -\frac{1}{2} \sqrt{6} - \frac{1}{2} \sqrt{10}\)

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

\({7 \over 2 \sqrt{5}} = {7 \over 2 \sqrt{5}} ⋅ {\sqrt{5} \over \sqrt{5}} = {7 \sqrt{5} \over 2 ⋅ 5} = \frac{7}{10} \sqrt{5} \text{.}\)