Tính
a) \(\sqrt {\frac{{16}}{{121}}} \)
b) \(\sqrt {4\frac{{21}}{{25}}} \)
c) \(\sqrt {\frac{{6,4}}{{8,1}}} \)
d) \(\frac{{\sqrt {300} }}{{\sqrt {27} }}\)
e) \(\frac{{\sqrt 6 }}{{\sqrt {150} }}\)
g) \(\sqrt {\frac{3}{2}} :\sqrt {\frac{1}{{24}}} \)
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Dựa vào: Với số thực a không âm và số thực b dương, ta có \(\sqrt {\frac{a}{b}} = \frac{{\sqrt a }}{{\sqrt b }}\).
a) \(\sqrt {\frac{{16}}{{121}}} = \frac{{\sqrt {16} }}{{\sqrt {121} }} = \frac{{\sqrt {{4^2}} }}{{\sqrt {{{11}^2}} }} = \frac{4}{{11}}.\)
b) \(\sqrt {4\frac{{21}}{{25}}} = \sqrt {\frac{{121}}{{25}}} = \frac{{\sqrt {{{11}^2}} }}{{\sqrt {{5^2}} }} = \frac{{11}}{5}.\)
c) \(\sqrt {\frac{{6,4}}{{8,1}}} = \sqrt {\frac{{64}}{{81}}} = \frac{{\sqrt {{8^2}} }}{{\sqrt {{9^2}} }} = \frac{8}{9}\).
d) \(\frac{{\sqrt {300} }}{{\sqrt {27} }} = \sqrt {\frac{{300}}{{27}}} = \sqrt {\frac{{100}}{9}} = \frac{{\sqrt {{{10}^2}} }}{{\sqrt {{3^2}} }} = \frac{{10}}{3}\) .
e) \(\frac{{\sqrt 6 }}{{\sqrt {150} }} = \sqrt {\frac{6}{{150}}} = \sqrt {\frac{1}{{25}}} = \frac{1}{{\sqrt {{5^2}} }} = \frac{1}{5}\).
g) \(\sqrt {\frac{3}{2}} :\sqrt {\frac{1}{{24}}} = \sqrt {\frac{3}{2}:\frac{1}{{24}}} = \sqrt {36} = \sqrt {{6^2}} = 6\).