a) (10 + 4x - 3x + x - 12x = 10 + (4 - 3 + 1 - 12)x = 10 - 10x)
б) $$\frac{2}{c-2} - \frac{4}{c-3} + 2(40+c) = \frac{2(c-3) - 4(c-2)}{(c-2)(c-3)} + 80 + 2c = \frac{2c - 6 - 4c + 8}{c^2 - 5c + 6} + 80 + 2c = \frac{-2c + 2}{c^2 - 5c + 6} + 80 + 2c$$
в) (4y + 3x - 10 - 6y + 6x + 15 = (4y - 6y) + (3x + 6x) + (-10 + 15) = -2y + 9x + 5)
Ответы:
a) (10 - 10x)
б) $$\frac{-2c + 2}{c^2 - 5c + 6} + 80 + 2c$$
в) (-2y + 9x + 5)