Решите уравнение: x^2-(11x+2)/6=0.

\[\ \ x² - \frac{11x + 2}{6} = 0\]

\[\frac{11x + 2}{6} = x^{2}\]

\[11x + 2 = 6x^{2}\]

\[6x^{2} - 11x - 2 = 0\]

\[D = b^{2} - 4ac = 121 - 4 \cdot 6 \cdot ( - 2) =\]

\[= 121 + 48 = 169\]

\[x_{1} = \frac{11 + 13}{12} = \frac{24}{12} = 2\]

\[x_{2} = \frac{11 - 13}{12} = - \frac{2}{12} = - \frac{1}{6}\]

\[Ответ:x = 2\ и\ \ x = - \frac{1}{6}.\]

\[x^{4} - 35x^{2} - 36 = 0\]

\[Пусть\ t = x^{2};\ \ t \geq 0\]

\[t^{2} - 35t - 36 = 0\]

\[t_{1} + t_{2} = 35\]

\[t_{1} \cdot t_{2} = - 36 \Longrightarrow t_{1} = 36\ \ \ и\ \ \ t_{2} = - 1\ \]

\[(не\ подходит)\]

\[x^{2} = 36\]

\[x = \pm 6\]

\[Ответ:x = \pm 6.\]

\[\frac{5x² + 3x - 2}{25x² - 4} = \frac{(5x - 2)(x + 1)}{(5x - 2)(5x + 2)} = \frac{x + 1}{5x + 2}\]

\[5x^{2} + 3x - 2 = 5 \cdot (x - 0,4)(x + 1) =\]

\[= (5x - 2)(x + 1)\]

\[D = b^{2} - 4ac = 9 - 4 \cdot 5 \cdot ( - 2) =\]

\[= 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + 7}{10} = \frac{4}{10} = 0,4\]

\[x_{2} = \frac{- 3 - 7}{10} = - \frac{10}{10} = - 1.\]

\[x^{2} + \text{px} + 72 = 0\ \ \ \ и\ \ \ x_{1} = - 9\]

\[x_{1} + x_{2} = - p \Longrightarrow - 9 - 8 = - p \Longrightarrow - 17 =\]

\[= - p \Longrightarrow p = 17.\]

\[x_{1} \cdot x_{2} = 72 \Longrightarrow - 9 \cdot x_{2} = 72 \Longrightarrow x_{2} = - 8.\]

\[Ответ:x_{2} = - 8\ \ \ и\ \ \ p = 17.\]

\[\ 6x² - 3x = 0\]

\[3x(2x - 1) = 0\]

\[x = 0\ \ \ \ \ \ \ \ \ 2x - 1 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{1}{2}\]

\[Ответ:\ \ x = 0\ \ и\ \ x = \frac{1}{2}.\]