Cho các số \(7;\,\,\sqrt {121} ;\,\,\frac{{67}}{3};\,\,\sqrt {88} ;\,\,4,32;\,\, - 2;\,\,\sqrt {12} ;\,\,\pi \). Trong đó,

Đáp án đúng là: D

Nhận thấy \(\sqrt 5 \) là số vô tỉ.

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 \(x = 0,11\).