Question

Решите уравнение

Answer 1

$\displaystyle \frac{x^2}{4} + \frac{9}{x^2}=3\left( \frac{x}{2}- \frac{3}{x} \right ) +1 \frac{3}{4} \ | \ \cdot \ 4x^2, \ x \ne 0\\\\ f(1) = 1 - 6 - 7 + 36 + 36 \ne 0, \ f(-1) = 1 + 6 -7 - 36 + 36 = 0\\\\ x^4 + 36 = 6x^3 - 36x + 7x^2, \ x^4 - 6x^3 - 7x^2 + 36x + 36 = 0\\\\ x^4 + x^3 - 7x^3 - 7x^2 + 36x + 36 = \\\\ = x^3(x + 1) - 7x^2(x + 1) + 36(x + 1) = \\\\ =(x^3 - 7x^2 + 36)(x + 1) = 0, \ x + 1 = 0, \ \boxed{x_1 = -1}$

$\displaystyle x^3 -7x^2 +36 = 0, \ 36 = 6 \cdot 6 = 3 \cdot 2 \cdot 3 \cdot 2 = 3^2 \cdot 2^2\\\\ f(2) = 8 - 28 + 36 \ne 0, \ f(-2) = -8 - 28 + 36 = 0\\\\ x^3 + 2x^2 - 9x^2 -18x + 18x + 36 = \\\\ = x^2(x + 2) - 9x(x + 2) + 18(x + 2) = \\\\ = (x^2 - 9x + 18)(x + 2) = 0, \ x + 2 = 0, \ \boxed{x_2 = -2}\\\\ x^2 - 9x + 18 = 0\\\\ x_3 + x_4 = 3 + 6, \ x_3 \cdot x_4 = 18 = 3 \cdot 6\\\\ \boxed{x_3 = 3, \ x_4 = 6}$

Первые корни уравнения найдены перебором (начиная с наименьших по абсолютной величине целых), оставшиеся два по формулам Виета.

$\displaystyle \frac{x^2}{4} + \frac{9}{x^2}=3\left( \frac{x}{2}- \frac{3}{x} \right ) +1 \frac{3}{4}, \ \frac{7}{4} = \frac{12 - 5}{4}\\\\ \left(\frac{x^2}{4} - \frac{12}{4} + \frac{9}{x^2}\right) - 3\left( \frac{x}{2}- \frac{3}{x} \right) + \frac{5}{4} = 0\\\\ \ \left( \frac{x}{2}- \frac{3}{x} \right)^2 - 3\left( \frac{x}{2}- \frac{3}{x} \right) +\frac{5}{4} = 0, \ t = \frac{x}{2}- \frac{3}{x}\ ; \ [\ x \ne 0 \ ]$

$\displaystyle t^2 - 3t + \frac{5}{4} = 0 \\\\ \text{D} = 9 - 5 = 4 \\\\ \ t_1 = \frac{3 - 2}{2} = 0.5, \ t_2 = \frac{3 + 2}{2} = 2.5\\\\ 1) \ \frac{x}{2}- \frac{3}{x} = \frac{1}{2} \ | \ \cdot \ 2x\\\\ x^2 - 6 = x, \ x^2 - x - 6 = 0\\\\ x_1 + x_2 = 1 = 3 - 2, \ x_1x_2 = -6 = (-2)\cdot 3\\\\ \boxed{x_1 = -2, \ x_2 = 3}$

$\displaystyle 2) \ \frac{x}{2}- \frac{3}{x} = \frac{5}{2} \ | \ \cdot \ 2x\\\\ x^2 - 6 = 5x, \ x^2 - 5x - 6 = 0\\\\ x_1 + x_2 = 5 = 6 - 1, \ x_1x_2 = -6 = (-1)\cdot 6\\\\ \boxed{x_3 = -1, \ x_4 = 6}$