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

Question

Алгебра

Боряха

16 сентября, 05:10

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

+5

Ответы (2)

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Виолетта · Answer 1

6x^2-x=y, 3y^2-4y+1=0, y1=1, y2=1/3,

6x^2-x-1=0, x1=1/2, x2=-1/3,

6x^2-x-1/3=0, x3=1/3, x4=-1/6;

Мариамна · Answer 2

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

Замена: 6 х^2 - х = t

3t^2 - 4t + 1 = 0

(t - 1) (3t - 1) = 0

t1 = 1; t2 = 1/3

6x^2 - x = 1 или 6 х^2 - х = 1/3 |*3

6x^2 - x - 1 = 0 или 18 х^2 - 3 х - 1 = 0

D = (-1) ^2 - 4*6 * (-1) ; D = (-3) ^2 - 4*18 * (-1)

D = 25 = 5^2; D = 81 = 9^2

x1 = 1 + 5 / 12; x2 = 1 - 5 / 12; x3 = 3 + 9 / 36; x4 = 3 - 9 / 36

x1 = 1/2; x2 = - 1/3; x3 = 1/3; x4 = - 1/6

Ответ: 1/2; - 1/3; 1/3; - 1/6