4x2+12x+12/x+4/x2=47
, нужно решить заменой ​

1. Преобразуем уравнение:

4х^2 + 12х + 12/х + 4/х^2 = 47;

4(х^2 + 2 + 1/x^2) - 8 + 12(х + 1/х) - 47 = 0;

4(х + 1/x)^2 + 12(х + 1/х) - 55 = 0.

  2. Замена:

х + 1/x = t;

4t^2 + 12t - 55 = 0;

D/4 = 6^2 + 4 * 55 = 36 + 220 = 256 = 16^2;

t = (-6 ± 16)/4;

t1 = (-6 - 16)/4 = -22/4 = -11/2;

t2 = (-6 + 16)/4 = 10/4 = 5/2.

  3. Обратная замена:

х + 1/x = t;

х^2 + 1 = tx;

х^2 - tx + 1 = 0;

  1) t = -11/2;

х^2 + 11/2 * x + 1 = 0;

2х^2 + 11x + 2 = 0;

D = 11^2 - 4 * 2 * 2 = 121 - 16 = 105;

x1/2 = (-11 ± √105)/4;

  2) t = 5/2;

х^2 - 5/2 * x + 1 = 0;

2х^2 - 5x + 2 = 0;

D = 5^2 - 4 * 2 * 2 = 25 - 16 = 9;

x = (5 ± √9)/4 = (5 ± 3)/4;

x3 = (5 - 3)/4 = 2/4 = 1/2;

x4 = (5 + 3)/4 = 8/4 = 2.

  ответ: (-11 ± √105)/4; 1/2; 2.