Решите уравнение (3х^2-x-2)^2+(6x^2+x-1)^4=0

(3 * х^2 x - 2)^2 + (6 * x^2 + x - 1)^4 = 0;

1) 3 * x^2 x 2 = 0;

D = b^2 4 * a * c = (-1)^2 4 * 3 * (-2) = 1 + 12 * 2 = 1 + 24 = 25 = 5^2;

x1 = (-(-1) + 5)/(2 * 3) = (1 + 5)/6 = 6/6 = 1;

x2 = (-(-1) 5)/(2 * 3) = (1 5)/6 = -4/6 = -2/3;

2) 6 * x^2 + x 1 = 0;

D = 1^2 4 * 6 * (-1) = 1 + 24 * 1 = 1 + 24 = 25 = 5^2;

x1 = (-1 + 5)/(2 * 6) = 4/12 = 1/3;

x2 = (-1 5)/(2 * 6) = -6/12 = -1/2;

Отсюда получаем:

(3 * (x 1/3) * (x + ))^2 + (6 * (x 1/3) * (x + ))^4 = 0;

3 * 3 * (x 1/3)^2* (x + )^2 + 6 * 6 * (x 1/3)^4 * (x + )^4 = 0;

(x 1/3)^2 * (x + )^2 * (9 + 36 * (x 1/3)^2 * (x + )^2) = 0;

(x 1/3)^2 * (x + )^2 * (9 + 3 * 3 * 2 * 2 * (x 1/3) * (x 1/3) * (x + ) * (x + )) = 0;

(x 1/3)^2 * (x + )^2 * (9 + (3 * x 1)^2 * (2 * x + 1)^2) = 0;

1) 9 + (3 * x 1)^2 * (2 * x + 1)^2 = 0;

Уравнение не имеет корней.

2) (x 1/3)^2 = 0;

x 1/3 = 0;

x = 1/3;

3) (x + )^2 = 0;

x + = 0;

x = -1/2;

Ответ: х = 1/3 и х = -1/2.