128. Найдите значение выражения (41/4x+31/3)-91/3-х, если х=13/3; х=2,4.

При $$x = 1\frac{1}{3} = \frac{4}{3}$$:

$$(4\frac{1}{4} \cdot \frac{4}{3} + 3\frac{1}{3}) - 9\frac{1}{3} = (\frac{17}{4} \cdot \frac{4}{3} + \frac{10}{3}) - \frac{28}{3} = (\frac{17}{3} + \frac{10}{3}) - \frac{28}{3} = \frac{27}{3} - \frac{28}{3} = \frac{-1}{3} = -\frac{1}{3}$$

При $$x = 2.4 = \frac{24}{10} = \frac{12}{5}$$:

$$(4\frac{1}{4} \cdot \frac{12}{5} + 3\frac{1}{3}) - 9\frac{1}{3} = (\frac{17}{4} \cdot \frac{12}{5} + \frac{10}{3}) - \frac{28}{3} = (\frac{17 \cdot 3}{5} + \frac{10}{3}) - \frac{28}{3} = (\frac{51}{5} + \frac{10}{3}) - \frac{28}{3} = (\frac{51 \cdot 3 + 10 \cdot 5}{15}) - \frac{28}{3} = \frac{153 + 50}{15} - \frac{28}{3} = \frac{203}{15} - \frac{28}{3} = \frac{203 - 28 \cdot 5}{15} = \frac{203 - 140}{15} = \frac{63}{15} = \frac{21}{5} = 4\frac{1}{5}$$

Ответ: $$\frac{-1}{3}; 4\frac{1}{5}$$