245. 1) (2\frac{3}{5}b - \frac{3}{4}b²) + (\frac{1}{4}b² - 1\frac{3}{5}b);

(2\frac{3}{5}b - \frac{3}{4}b^2) + (\frac{1}{4}b^2 - 1\frac{3}{5}b) = 2\frac{3}{5}b - \frac{3}{4}b^2 + \frac{1}{4}b^2 - 1\frac{3}{5}b = (2\frac{3}{5} - 1\frac{3}{5})b + (-\frac{3}{4} + \frac{1}{4})b^2 = (2\frac{3}{5} - 1\frac{3}{5})b + (\frac{-3 + 1}{4})b^2 = 1b - \frac{2}{4}b^2 = b - \frac{1}{2}b^2 Ответ: b - \frac{1}{2}b^2