Ответ:
а) 7x^2-x=0 ---------> x*(7x-1) = 0 -----> x1=0; x2=1\7
б) (6-2x)^2=3x-9
[2*(3-x)]^2 = 3*(x-3)
4 * (3-x)^2 = (-3) * (3-x)
4*(3-x)^2 + 3*(3-x) = 0
(3-x) * [4*(3-x) + 3] = 0
(3-x) * (15 - 4x) = 0
(3-x) = 0 ------------> x1 = 3
(15-4x) = 0 ---------> x2 = 15\4
в) 2x^3 - 8x^2 + 5x - 20 = 0
2x^2 * (x-4) + 5*(x-4) = 0
(x-4) * (2x^2 + 5) = 0
(x-4) = 0 --------------------> x1 = 4
(2x^2 + 5) > 0 ------------------> при любых х
Объяснение: