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

Приведем подобные члены многочлена: $$\frac{2}{3} ab^2 + \frac{1}{7} a^2b + \frac{1}{6} ba^2 - \frac{4}{7} ba^2 = \frac{2}{3}ab^2 + (\frac{1}{7} + \frac{1}{6} - \frac{4}{7})a^2b = \frac{2}{3}ab^2 + (\frac{6 + 7 - 24}{42})a^2b = \frac{2}{3}ab^2 + (\frac{13 - 24}{42})a^2b = \frac{2}{3}ab^2 - \frac{11}{42}a^2b$$.

Ответ: $$\frac{2}{3}ab^2 - \frac{11}{42}a^2b$$