Решите уравнение: (x^2-7x)/8-1=0.

\[\ \frac{x^{2} - 7x}{8} - 1 = 0\]

\[\frac{x^{2} - 7x}{8} = 1\]

\[x^{2} - 7x = 8\]

\[x^{2} - 7x - 8 = 0\]

\[x_{1} + x_{2} = 7\]

\[x_{1} \cdot x_{2} = - 8 \Longrightarrow x_{1} = 8\ \ \ и\ \ x_{2} = - 1\]

\[Ответ:\ \ x_{1} = 8\ \ \ и\ x_{2} = - 1.\]

\[x^{4} - 13x^{2} + 36 = 0\]

\[Пусть\ \ \ \ t = x^{2},\ \ \ \ t \geq 0\]

\[t^{2} - 13t + 36 = 0\]

\[t_{1}{+ t}_{2} = 13\]

\[t_{1} \cdot t_{2} = 36 \Longrightarrow t_{1} = 9\ и\ t_{2} = 4.\]

\[x^{2} = 9\ \ \ \ \ \ \ \ \ \ \ \ x^{2} = 4\]

\[x = \pm 3\ \ \ \ \ \ \ \ \ \ x = \pm 2\]

\[Ответ:x = \pm 3\ \ и\ \ x = \pm 2.\]

\[\frac{6x^{2} - x - 1}{9x^{2} - 1} = \frac{(2x - 1)(3x + 1)}{(3x - 1)(3x + 1)} = \frac{2x - 1}{3x - 1}\]

\[6x^{2} - x - 1 = 6 \cdot \left( x - \frac{1}{2} \right)\left( x + \frac{1}{3} \right) =\]

\[= (2x - 1)(3x + 1)\]

\[D = b^{2} - 4ac = 1 - 4 \cdot 6 \cdot ( - 1) =\]

\[= 1 + 24 = 25\]

\[x_{1} = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}\]

\[x_{2} = \frac{1 - 5}{12} = \frac{- 4}{12} = - \frac{1}{3}.\]

\[x^{2} + \text{kx} + 45 = 0\ \ \ и\ \ \ x_{1} = 5\]

\[x_{1} + x_{2} = - k \Longrightarrow 5 + 9 = - k \Longrightarrow 14 =\]

\[= - k \Longrightarrow k = - 14.\]

\[x_{1} \cdot x_{2} = 45 \Longrightarrow 5 \cdot x_{2} = 45 \Longrightarrow x_{2} = 9.\]

\[Ответ:\ \ x_{2} = 9\ \ \ и\ \ k = - 14.\]

\[\ 6x - 5x^{2} = 0\]

\[x(6 - 5x) = 0\]

\[x = 0\ \ \ \ \ \ \ \ \ \ 6 - 5x = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 5x = - 6\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1,2\]

\[Ответ:x = 0\ \ \ и\ \ \ x = 1,2.\]