Упростите:
а) $(2a^2b - 10b^3) - (4a^2b - 12b^3)$;
б) $(3xy^2 + 7x^2y) - (2xy^2 - 6x^2y)$;
в) 12ab − 30bc − 3cx − (15bc + 9cx);
г) (10abc − 8bcx − 21cxy) − (−6abc + bcx − cxy);
д) (0,6ab − 0,5bc + cx) − (2,5bc − 0,5ab − cx);
е) $(\frac{1}{2}x^2y^2 - \frac{2}{3}ab - \frac{5}{6}a^2b - 1) - (a^2b - \frac{1}{3}x^2y^2 + \frac{1}{12}ab - \frac{1}{4})$.
$(2a^2b - 10b^3) - (4a^2b - 12b^3) = 2a^2b - 10b^3 - 4a^2b + 12b^3 = -2a^2b + 2b^3$
$(3xy^2 + 7x^2y) - (2xy^2 - 6x^2y) = 3xy^2 + 7x^2y - 2xy^2 + 6x^2y = xy^2 + 13x^2y$
12ab − 30bc − 3cx − (15bc + 9cx) = 12ab − 30bc − 3cx − 15bc − 9cx = 12ab − 45bc − 12cx
(10abc − 8bcx − 21cxy) − (−6abc + bcx − cxy) = 10abc − 8bcx − 21cxy + 6abc − bcx + cxy = 16abc − 9bcx − 20cxy
(0,6ab − 0,5bc + cx) − (2,5bc − 0,5ab − cx) = 0,6ab − 0,5bc + cx − 2,5bc + 0,5ab + cx = 1,1ab − 3bc + 2cx
$(\frac{1}{2}x^2y^2 - \frac{2}{3}ab - \frac{5}{6}a^2b - 1) - (a^2b - \frac{1}{3}x^2y^2 + \frac{1}{12}ab - \frac{1}{4}) = \frac{3}{6}x^2y^2 - \frac{8}{12}ab - \frac{5}{6}a^2b - 1 - a^2b + \frac{2}{6}x^2y^2 - \frac{1}{12}ab + \frac{1}{4} = \frac{5}{6}x^2y^2 - \frac{9}{12}ab - 1\frac{5}{6}a^2b - \frac{3}{4} = \frac{5}{6}x^2y^2 - \frac{3}{4}ab - 1\frac{5}{6}a^2b - \frac{3}{4}$
Пожауйста, оцените решение
