Решите ур-ие, используя введение новой переменной. a) (x^2+x-1) (x^2+x+2) = 40 c) (2x^2+x-1) (2x^2+x-4) + 2=0

a) Обозначим:

x^2 + x - 1 = y; (x^2 + x - 1) (x^2 + x + 2) = 40; y (y + 3) = 40; y^2 + 3y - 40 = 0; y = (-3 ± √ (9 + 160)) / 2 = (-3 ± 13) / 2;

1) y = (-3 - 13) / 2 = - 8;

x^2 + x - 1 = - 8; x^2 + x + 7 = 0; D = 1^2 - 4 * 7 = 1 - 28 = - 27 < 0, нет решения.

2) y2 = (-3 + 13) / 2 = 5;

x^2 + x - 1 = 5; x^2 + x - 6 = 0; x = (-1 ± √ (1 + 24)) / 2 = (-1 ± 5) / 2; x1 = (-1 - 5) / 2 = - 3; x2 = (-1 + 5) / 2 = 2.

c) Обозначим:

2x^2 + x - 1 = y; (2x^2 + x - 1) (2x^2 + x - 4) + 2 = 0; y (y - 3) + 2 = 0; y^2 - 3y + 2 = 0; y1 = 1; y2 = 2;

1) y = 1;

2x^2 + x - 1 = 1; 2x^2 + x - 2 = 0; D = 1 + 4 * 2 * 2 = 17; x1/2 = (-1 ± √17) / 4.

2) y = 2;

2x^2 + x - 1 = 2; 2x^2 + x - 3 = 0; D = 1 + 4 * 2 * 3 = 25; x = (-1 ± 5) / 4; x3 = - 3/2; x4 = 1.

Ответ:

a) - 3; 2; c) (-1 ± √17) / 4; - 3/2; 1.