10. Решите уравнение 6 (x²+1/x²) + 5 (x+1/x) – 38 = 0.

Пусть $$ t = x + \frac{1}{x} $$. Тогда $$ t^2 = x^2 + 2 + \frac{1}{x^2} $$, следовательно, $$ x^2 + \frac{1}{x^2} = t^2 - 2 $$.

Подставим в исходное уравнение:

$$ 6(t^2 - 2) + 5t - 38 = 0 $$ $$ 6t^2 - 12 + 5t - 38 = 0 $$ $$ 6t^2 + 5t - 50 = 0 $$ Решим квадратное уравнение относительно t:

$$ D = 5^2 - 4 \cdot 6 \cdot (-50) = 25 + 1200 = 1225 $$ $$ t_1 = \frac{-5 + \sqrt{1225}}{2 \cdot 6} = \frac{-5 + 35}{12} = \frac{30}{12} = \frac{5}{2} $$ $$ t_2 = \frac{-5 - \sqrt{1225}}{2 \cdot 6} = \frac{-5 - 35}{12} = \frac{-40}{12} = -\frac{10}{3} $$ Теперь вернемся к замене $$ t = x + \frac{1}{x} $$:

1) $$ x + \frac{1}{x} = \frac{5}{2} $$ $$ \frac{x^2 + 1}{x} = \frac{5}{2} $$ $$ 2(x^2 + 1) = 5x $$ $$ 2x^2 - 5x + 2 = 0 $$ $$ D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 $$ $$ x_1 = \frac{5 + \sqrt{9}}{2 \cdot 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2 $$ $$ x_2 = \frac{5 - \sqrt{9}}{2 \cdot 2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} $$ 2) $$ x + \frac{1}{x} = -\frac{10}{3} $$ $$ \frac{x^2 + 1}{x} = -\frac{10}{3} $$ $$ 3(x^2 + 1) = -10x $$ $$ 3x^2 + 10x + 3 = 0 $$ $$ D = 10^2 - 4 \cdot 3 \cdot 3 = 100 - 36 = 64 $$ $$ x_3 = \frac{-10 + \sqrt{64}}{2 \cdot 3} = \frac{-10 + 8}{6} = \frac{-2}{6} = -\frac{1}{3} $$ $$ x_4 = \frac{-10 - \sqrt{64}}{2 \cdot 3} = \frac{-10 - 8}{6} = \frac{-18}{6} = -3 $$

Ответ: 2, 1/2, -1/3, -3