8) \frac{1}{5}a2b2+\frac{2}{5}a2b+\frac{3}{5}ab2 + \frac{4}{5}a2b²=

Приведем подобные члены многочлена: $$\frac{1}{5}a^2b^2 + \frac{2}{5}a^2b + \frac{3}{5}ab^2 + \frac{4}{5}a^2b^2 = (\frac{1}{5}a^2b^2 + \frac{4}{5}a^2b^2) + \frac{2}{5}a^2b + \frac{3}{5}ab^2 = \frac{5}{5}a^2b^2 + \frac{2}{5}a^2b + \frac{3}{5}ab^2 = a^2b^2 + \frac{2}{5}a^2b + \frac{3}{5}ab^2$$

Ответ: $$a^2b^2 + \frac{2}{5}a^2b + \frac{3}{5}ab^2$$