Найдите корни уравнения: 2(x^2+1/x^2)+x-1/x=7.

Ответ:

\[2 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) + x - \frac{1}{x} = 7\]

\[x^{2} + \frac{1}{x^{2}} = x^{2} + \frac{1}{x^{2}} - 2 + 2 =\]

\[= \left( x - \frac{1}{x} \right)^{2} + 2\]

\[Пусть\ y = x - \frac{1}{x}:\]

\[2 \cdot \left( y^{2} + 2 \right) + y - 7 = 0\]

\[2y^{2} + 4 + y - 7 = 0\]

\[2y^{2} + y - 3 = 0\]

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 + 5}{4} = 1;\ \ \ \ \]

\[y_{2} = \frac{- 1 - 5}{4} = - 1,5\]

\[Подставим:\]

\[1)\ x^{\backslash x} - \frac{1}{x} = 1^{\backslash x}\]

\[x^{2} - x - 1 = 0\]

\[D = 1 + 4 = 5\]

\[x_{1,2} = \frac{1 \pm \sqrt{5}}{2}.\]

\[2)\ x - \frac{1}{x} = - 1,5\ \ \ \ \ | \cdot 2x\]

\[2x^{2} + 3x - 2 = 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 + 5}{4} = 0,5;\ \]

\[\text{\ \ }x_{2} = \frac{- 3 - 5}{4} = - 2.\]

\[Ответ:x = - 2;\ \ x = 0,5;\ \]

\[\ x = \frac{1 \pm \sqrt{5}}{2}.\]