Tính: \(\begin{array}{l}a)\,\,\sqrt {\frac{{289}}{{225}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\sqrt {2\frac{{14}}{{25}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)\,\,\sqrt {\frac{{0,25}}{9}} \\d)\,\,\sqrt {1\frac{{16}}{9}.5\frac{4}{9}.0,01} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e)\,\,\sqrt {\frac{{{{149}^2} – {{76}^2}}}{{{{457}^2} – {{384}^2}}}} \end{array}\) A \(\begin{array}{l} a)\,\,\,\frac{{27}}{{15}} & & & b)\,\,\frac{8}{5}\\ c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\ e)\,\,\frac{{15}}{{29}} \end{array}\) B \(\begin{array}{l} a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\ c)\,\,\frac{1}{6} & & & d)\,\,\frac{17}{{24}}\\ e)\,\,\frac{{15}}{{29}} \end{array}\) C \(\begin{array}{l} a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\ c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\ e)\,\,\frac{{25}}{{29}} \end{array}\) D \(\begin{array}{l} a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\ c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\ e)\,\,\frac{{15}}{{29}} \end{array}\)

Tính:

\(\begin{array}{l}a)\,\,\sqrt {\frac{{289}}{{225}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\sqrt {2\frac{{14}}{{25}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)\,\,\sqrt {\frac{{0,25}}{9}} \\d)\,\,\sqrt {1\frac{{16}}{9}.5\frac{4}{9}.0,01} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e)\,\,\sqrt {\frac{{{{149}^2} – {{76}^2}}}{{{{457}^2} – {{384}^2}}}} \end{array}\)

A \(\begin{array}{l}

a)\,\,\,\frac{{27}}{{15}} & & & b)\,\,\frac{8}{5}\\

c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\

e)\,\,\frac{{15}}{{29}}

\end{array}\)

B \(\begin{array}{l}

a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\

c)\,\,\frac{1}{6} & & & d)\,\,\frac{17}{{24}}\\

e)\,\,\frac{{15}}{{29}}

\end{array}\)

C \(\begin{array}{l}

a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\

c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\

e)\,\,\frac{{25}}{{29}}

\end{array}\)

D \(\begin{array}{l}

a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\

c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\

e)\,\,\frac{{15}}{{29}}

\end{array}\)

Hướng dẫn Chọn đáp án là: D

Lời giải chi tiết:

\(\begin{array}{l}a)\,\,\,\sqrt {\frac{{289}}{{225}}} = \frac{{\sqrt {289} }}{{\sqrt {225} }} = \frac{{\sqrt {{{17}^2}} }}{{\sqrt {{{15}^2}} }} = \frac{{17}}{{15}}.\\b)\,\,\sqrt {2\frac{{14}}{{25}}} = \sqrt {\frac{{64}}{{25}}} = \frac{{\sqrt {64} }}{{\sqrt {25} }} = \frac{{\sqrt {{8^2}} }}{{\sqrt {{5^2}} }} = \frac{8}{5}.\\c)\,\,\sqrt {\frac{{0,25}}{9}} = \frac{{\sqrt {0,25} }}{{\sqrt 9 }} = \frac{{\sqrt {0,{5^2}} }}{{\sqrt {{3^2}} }} = \frac{{0,5}}{3} = \frac{1}{6}.\\d)\,\,\sqrt {1\frac{9}{{16}}.5\frac{4}{9}.0,01} = \sqrt {\frac{{25}}{{16}}.\frac{{49}}{9}.\frac{1}{{100}}} \\ = \sqrt {\frac{{25}}{{16}}} .\sqrt {\frac{{49}}{9}} .\sqrt {\frac{1}{{100}}} = \sqrt {\frac{{{5^2}}}{{{4^2}}}} .\sqrt {\frac{{{7^2}}}{{{3^2}}}} .\sqrt {\frac{1}{{{{10}^2}}}} = \frac{5}{4}.\frac{7}{3}.\frac{1}{{10}} = \frac{7}{{24}}.\\e)\,\,\sqrt {\frac{{{{149}^2} – {{76}^2}}}{{{{457}^2} – {{384}^2}}}} = \sqrt {\frac{{\left( {149 – 76} \right)\left( {149 + 76} \right)}}{{\left( {457 – 384} \right)\left( {457 + 384} \right)}}} \\ = \sqrt {\frac{{73.225}}{{73.841}}} = \sqrt {\frac{{225}}{{841}}} = \frac{{\sqrt {225} }}{{\sqrt {841} }} = \frac{{15}}{{29}}.\end{array}\)