Ta có:
+) \(\displaystyle \sqrt {{{16} \over {25}}} = \sqrt {{{\left( {{4 \over 5}} \right)}^2}} = {4 \over 5}\)
+) \(\displaystyle {{\sqrt {16} } \over {\sqrt {25} }} = {4 \over 5}\)
\( \displaystyle \Rightarrow \sqrt {{{16} \over {25}}} = {{\sqrt {16} } \over {\sqrt {25} }}\)
