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

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

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