10) \frac{2}{3} ab² + \frac{1}{6} a²b + \frac{1}{7} b²a - \frac{4}{7} ba² =

Приведем подобные члены:

$$\frac{2}{3}ab^2 + \frac{1}{6}a^2b + \frac{1}{7}b^2a - \frac{4}{7}ba^2 = \frac{2}{3}ab^2 + \frac{1}{7}ab^2 + \frac{1}{6}a^2b - \frac{4}{7}a^2b = \frac{14}{21}ab^2 + \frac{3}{21}ab^2 + \frac{7}{42}a^2b - \frac{24}{42}a^2b = \frac{17}{21}ab^2 - \frac{17}{42}a^2b$$

Ответ: $$\frac{17}{21}ab^2 - \frac{17}{42}a^2b$$