Г) \( 3\frac{1}{3}-0,8-2\frac{3}{4}+0,3+1\frac{1}{12} \)

Решение:

\( 3\frac{1}{3}-0,8-2\frac{3}{4}+0,3+1\frac{1}{12} = 3\frac{1}{3} - \frac{8}{10} - 2\frac{3}{4} + \frac{3}{10} + 1\frac{1}{12} = 3\frac{1}{3} - \frac{4}{5} - 2\frac{3}{4} + \frac{3}{10} + 1\frac{1}{12} = \left(3\frac{1}{3}+1\frac{1}{12}\right) - \left(\frac{4}{5}+2\frac{3}{4}-\frac{3}{10}\right) = \left(3\frac{4}{12}+1\frac{1}{12}\right) - \left(\frac{8}{10}+2\frac{3}{4}-\frac{3}{10}\right) = 4\frac{5}{12} - \left(\frac{5}{10}+2\frac{3}{4}\right) = 4\frac{5}{12} - \left(\frac{1}{2}+2\frac{3}{4}\right) = 4\frac{5}{12} - \left(\frac{2}{4}+2\frac{3}{4}\right) = 4\frac{5}{12} - 2\frac{5}{4} = 4\frac{5}{12} - 2\frac{15}{12} = 4\frac{5}{12} - 3\frac{3}{12} = 1\frac{2}{12} = 1\frac{1}{6} \)

Ответ: 1\( \frac{1}{6} \)