Ta có: \(\sqrt {\frac{{16}}{{25}}} = \sqrt {{{\left( {\frac{4}{5}} \right)}^2}} = \frac{4}{5}\)
\(\frac{{\sqrt {16} }}{{\sqrt {25} }} = \frac{{\sqrt {4{}^2} }}{{\sqrt {{5^2}} }} = \frac{4}{5}\).
Vậy \(\sqrt {\frac{{16}}{{25}}} = \frac{{\sqrt {16} }}{{\sqrt {25} }}\).
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