\(\sqrt{700} = \sqrt{100} ⋅ \sqrt{7} = 10 \sqrt{7} \text{.}\)

\(7 \sqrt{175} = 7 ⋅ \sqrt{25} ⋅ \sqrt{7} = 7 ⋅ 5 ⋅ \sqrt{7} = 35 \sqrt{7} \text{.}\)

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

\({30 \sqrt{64} \over 5 \sqrt{8}} = {30 \over 5} ⋅ {\sqrt{64} \over \sqrt{8}} = 6 \sqrt{8} = 6 ⋅ \sqrt{4} ⋅ \sqrt{2} = 6 ⋅ 2 ⋅ \sqrt{2} = 12 \sqrt{2}\)

\(\sqrt{\frac{40}{49}} = {\sqrt{40} \over \sqrt{49}} = {\sqrt{40} \over 7} = \frac{1}{7} \sqrt{40} = \frac{1}{7} ⋅ 2 ⋅ \sqrt{10} = \frac{2}{7} \sqrt{10} \text{.}\)

\(\sqrt{300} + \sqrt{12} = \sqrt{100} ⋅ \sqrt{3} + \sqrt{4} ⋅ \sqrt{3} = 10 \sqrt{3} + 2 \sqrt{3} \text{.}\)

\(10 \sqrt{3} + 2 \sqrt{3} = 12 \sqrt{3} \text{.}\)

\(4 \sqrt{500} - 3 \sqrt{20} = 4 ⋅ \sqrt{100} ⋅ \sqrt{5} - 3 ⋅ \sqrt{4} ⋅ \sqrt{5} \text{.}\)

\(4 ⋅ 10 ⋅ \sqrt{5} - 3 ⋅ 2 ⋅ \sqrt{5} = 40 \sqrt{5} - 6 \sqrt{5} = 34 \sqrt{5} \text{.}\)

\({9 \over 4 \sqrt{5}} = {9 \over 4 \sqrt{5}} ⋅ {\sqrt{5} \over \sqrt{5}} = {9 \sqrt{5} \over 4 ⋅ 5} = \frac{9}{20} \sqrt{5} \text{.}\)

\(\sqrt{\frac{1}{44}} = {\sqrt{1} \over \sqrt{44}} = {1 \over \sqrt{44}} ⋅ {\sqrt{44} \over \sqrt{44}} = {\sqrt{44} \over 44} = \frac{1}{44} \sqrt{44} = \frac{1}{44} ⋅ 2 ⋅ \sqrt{11} = \frac{1}{22} \sqrt{11} \text{.}\)

\(\sqrt{4\frac{2}{3}} = \sqrt{\frac{14}{3}} = {\sqrt{14} \over \sqrt{3}} ⋅ {\sqrt{3} \over \sqrt{3}} = {\sqrt{42} \over 3} = \frac{1}{3} \sqrt{42} \text{.}\)

\(2 \sqrt{6} ⋅ 4 \sqrt{15} = 8 \sqrt{90} = 8 ⋅ \sqrt{9} ⋅ \sqrt{10} = 8 ⋅ 3 ⋅ \sqrt{10} = 24 \sqrt{10}\)