Решите уравнение методом замены переменной: (x^2-x-1)/x-6x/(x^2-x-1)=5.

Ответ:

\[ОДЗ:\ x^{2} - x - 1 \neq 0\]

\[D = ( - 1)^{2} - 4 \cdot 1 \cdot ( - 1) =\]

\[= 1 + 4 = 5\]

\[x_{1} \neq \frac{1 + \sqrt{5}}{2};\ \ x_{2} \neq \frac{1 - \sqrt{5}}{2}.\]

\[\left( x^{4} - 2x^{2} + 1 \right) - 7x\left( x^{2} - 1 \right) = 0\]

\[\left( x^{2} - 1 \right)^{2} - 7x\left( x^{2} - 1 \right) = 0\]

\[\left( x^{2} - 1 \right)\left( x^{2} - 1 - 7x \right) = 0\]

\[(x - 1)(x + 1)(x² - 7x - 1) = 0\]

\[x = 1;\ \ \ x = - 1.\ \ \]

\[x^{2} - 7x - 1 = 0\]

\[D = ( - 7)^{2} - 4 \cdot 1 \cdot ( - 1) =\]

\[= 49 + 4 = 53\]

\[x_{1} = \frac{7 + \sqrt{53}}{2};\ \ \ \ \ x_{2} = \frac{7 - \sqrt{53}}{2}.\]

\[Ответ:x = \pm 1;x = \frac{7 \pm \sqrt{53}}{2}\text{\ .}\]