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

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

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