Найдите сумму и разность многочленов:
1) $2,8b - 0,75b^2$ и $\frac{1}{4}b^2 - 1\frac{4}{5}b$;
2) $1\frac{2}{7}x^2 + 2\frac{4}{9}y$ и $2\frac{3}{14}x^2 - 1\frac{1}{6}y$.
$2,8b - 0,75b^2 + \frac{1}{4}b^2 - 1\frac{4}{5}b = (2,8b - 1,8b) + (-0,75b^2 + 0,25b^2) = b - 0,5b^2$
$2,8b - 0,75b^2 - (\frac{1}{4}b^2 - 1\frac{4}{5}b) = 2,8b - 0,75b^2 - 0,25b^2 + 1,8b = (2,8b + 1,8b) + (-0,75b^2 - 0,25b^2) = 4,6b - b^2$
$1\frac{2}{7}x^2 + 2\frac{4}{9}y + 2\frac{3}{14}x^2 - 1\frac{1}{6}y = (1\frac{2}{7}x^2 + 2\frac{3}{14}x^2) + (2\frac{4}{9}y - 1\frac{1}{6}y) = (1\frac{4}{14}x^2 + 2\frac{3}{14}x^2) + (2\frac{8}{18}y - 1\frac{3}{18}y) = 3\frac{7}{14}x^2 + 1\frac{5}{18}y = 3\frac{1}{2}x^2 + 1\frac{5}{18}y$
$1\frac{2}{7}x^2 + 2\frac{4}{9}y - (2\frac{3}{14}x^2 - 1\frac{1}{6}y) = 1\frac{2}{7}x^2 + 2\frac{4}{9}y - 2\frac{3}{14}x^2 + 1\frac{1}{6}y = (1\frac{2}{7}x^2 - 2\frac{3}{14}x^2) + (2\frac{4}{9}y + 1\frac{1}{6}y) = (1\frac{4}{14}x^2 - 1\frac{17}{14}x^2) + (2\frac{8}{18}y + 1\frac{3}{18}y) = -\frac{13}{14}x^2 + 3\frac{11}{18}y$
Пожауйста, оцените решение
