A. \( \frac{{133}}{{11}}\)
B. \( \frac{{403}}{{36}}\)
C. \( \frac{{12}}{{133}}\)
D. \( \frac{{133}}{{12}}\)
D
Ta có:
\(\begin{array}{l} 0,\left( {18} \right) = \frac{{18}}{{99}} = \frac{2}{{11}};\\ 2,0\left( {15} \right) = 2 + 0,0\left( {15} \right) = 2 + \frac{{15}}{{990}} = 2 + \frac{1}{{66}} = \frac{{133}}{{66}}\\ \to 0,\left( {18} \right).x = 2,0\left( {15} \right) \Leftrightarrow \frac{2}{{11}}.x = \frac{{133}}{{66}} \Leftrightarrow x = \frac{{133}}{{66}}:\frac{2}{{11}} \Leftrightarrow x = \frac{{133}}{{12}} \end{array}\)
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