\(\sqrt{0} = 0\)

\(\sqrt{36} = 6\)

\(\sqrt{121} = 11\)

\(\sqrt{4} + \sqrt{36} = 2 + 6 = 8\)

\(7 \sqrt{121} - 8 \sqrt{9} = 7 ⋅ 11 - 8 ⋅ 3 = 53\)

\(\sqrt{9} ⋅ \sqrt{36} = 3 ⋅ 6 = 18\)

\((6 \sqrt{3})^{2} = 6^{2} ⋅ \sqrt{3}^{2} = 36 ⋅ 3 = 108 \text{.}\)

\((6 \sqrt{3} - 2) (6 \sqrt{3} + 2) = (6 \sqrt{3})^{2} - 2^{2} = 36 ⋅ 3 - 4 = 108 - 4 = 104 \text{.}\)

\((\sqrt{3})^{2} = 3 \text{.}\)

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

\(8 \sqrt{6} + 2 \sqrt{6} = 10 \sqrt{6}\)

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

\({\sqrt{12} \over \sqrt{2}} = \sqrt{6}\)

\({48 \sqrt{48} \over 8 \sqrt{8}} = {48 \over 8} ⋅ {\sqrt{48} \over \sqrt{8}} = 6 \sqrt{6}\)

\({35 \sqrt{40} \over 7 \sqrt{10}} = {35 \over 7} ⋅ {\sqrt{40} \over \sqrt{10}} = 5 ⋅ \sqrt{4} = 5 ⋅ 2 = 10\)