\(1 \frac{3}{25} \cdot 2 \frac{1}{7} = \frac{1 25 + 3}{25} \cdot \frac{2 7 + 1}{7} = \frac{28}{25} \cdot \frac{15}{7}\)
\(\frac{28}{25} \cdot \frac{15}{7} = \frac{\cancel{28}^4}{\cancel{25}^5} \cdot \frac{\cancel{15}^3}{\cancel{7}^1} = \frac{4 3}{5 \u0002 1} = \frac{12}{5}\)
\(2 \frac{1}{9} \cdot \frac{27}{190} = \frac{2 9 + 1}{9} \cdot \frac{27}{190} = \frac{19}{9} \cdot \frac{27}{190}\)
\(\frac{19}{9} \cdot \frac{27}{190} = \frac{\cancel{19}^1}{\cancel{9}^1} \cdot \frac{\cancel{27}^3}{\cancel{190}^{10}} = \frac{1 3}{1 \u0002 10} = \frac{3}{10}\)
\(\frac{12}{5} - \frac{3}{10}\)
\(\frac{12 \u0002 2}{5 \u0002 2} - \frac{3}{10} = \frac{24}{10} - \frac{3}{10} = \frac{24 - 3}{10} = \frac{21}{10}\)
\(\frac{21}{10} = 2 \frac{1}{10}\)
Ответ: 2 1/10