Cho các biểu thức sau: \(A = \sqrt {9 \cdot 16} ;\,\,B = \sqrt 9  \cdot \sqrt {16} ;\,\,C = \sqrt {4 + 81} ;\,\,D = \sqrt 4  + \sqrt {81} \). Khi đó

Đáp án đúng là: D

Ta có: \(\sqrt {0,04}  = \sqrt {{{\left( {0,2} \right)}^2}}  = 0,2.\)

Đáp án: 0,11

Ta có: \(1\frac{4}{9} - \frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \sqrt {1\frac{7}{9}} \)

\(\frac{{13}}{9} - \frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \sqrt {\frac{{16}}{9}} \)

\(\frac{{13}}{9} - \frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \sqrt {{{\left( {\frac{4}{3}} \right)}^2}} \)

\(\frac{{13}}{9} - \frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \frac{4}{3}\)

\(\frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \frac{{13}}{9} - \frac{4}{3}\)

\(\frac{1}{{25}}:\left( {x + \frac{1}{4}} \right) = \frac{1}{9}\)

\(x + \frac{1}{4} = \frac{1}{{25}}:\frac{1}{9}\)

\(x + \frac{1}{4} = \frac{9}{{25}}\)

\(x = \frac{9}{{25}} - \frac{1}{4}\)

\(x = \frac{{11}}{{100}}\)

\(x = 0,11\).

Vậy \(x = 0,11\).