10) 2/3 ab² + 1/7 a²b+ 1/6 b²a – 4/7 ba² =

Приведем подобные члены:

$$\frac{2}{3}ab^2 + \frac{1}{7}a^2b + \frac{1}{6}b^2a - \frac{4}{7}ba^2 = \frac{2}{3}ab^2 + \frac{1}{6}ab^2 + \frac{1}{7}a^2b - \frac{4}{7}a^2b = (\frac{2}{3} + \frac{1}{6})ab^2 + (\frac{1}{7} - \frac{4}{7})a^2b = (\frac{4}{6} + \frac{1}{6})ab^2 + (\frac{-3}{7})a^2b = \frac{5}{6}ab^2 - \frac{3}{7}a^2b$$

Ответ: $$\frac{5}{6}ab^2 - \frac{3}{7}a^2b$$