50 баллов !область определения выражения

1) (3x^2 - x - 4)(3x^2 - x + 2) - 7 = 0 замена 3x^2 - x = y (y - 4)(y + 2) - 7 = 0 y^2 - 4y + 2y - 8 - 7 = y^2 - 2y - 15 = 0 (y - 5)(y + 3) = 0 а) y - 5 = 3x^2 - x - 5 = 0 d = 1 - 4*3(-5) = 1 + 60 = 61 x1 = (1 - √61)/6; x2 = (1 + √61)/6 б) y + 3 = 3x^2 - x + 3 = 0 d = 1 - 4*3*3 = 1 - 36 = -35 < 0 корней нет. ответ: x1 = (1 - √61)/6; x2 = (1 + √61)/6 2) (x^2 - 3x + 2)(x^2 - 7x + 12) = 4 (x - 1)(x - 2)(x - 3)(x - 4) = 4 (x - 1)(x - 4)*(x - 2)(x - 3) = 4 (x^2 - 5x + 4)(x^2 - 5x + 6) = 4 замена x^2 - 5x + 4 = y y(y + 2) = 4 y^2 + 2y - 4 = 0 d = 4 - 4*1(-4) = 4 + 16 = 20 = (2√5)^2 а) y1 = x^2 - 5x + 4 = (-2 - 2√5)/2 = -1 - √5 x^2 - 5x + 5 + √5 = 0 d = 25 - 4*(5 + √5) = 5 - 4√5 < 0 корней нет б) y2 = x^2 - 5x + 4 = -1 + √5 x^2 - 5x + 5 - √5 = 0 d = 25 - 4(5 - √5) = 5 + 4√5 x1 = (5 - √(5 + 4√5))/2; x2 = (5 + √(5 + 4√5))/2