Решите уравнение: x^2-(9x+2)/5=0.

\[\ x² - \frac{9x + 2}{5} = 0\]

\[x² = \frac{9x + 2}{5}\]

\[5x^{2} = 9x + 2\]

\[5x^{2} - 9x - 2 = 0\]

\[D = b^{2} - 4ac = 81 - 4 \cdot 5 \cdot ( - 2) =\]

\[= 81 + 40 = 121\]

\[x_{1} = \frac{9 - 11}{10} = - \frac{2}{10} = - 0,2\]

\[x_{2} = \frac{9 + 11}{10} = \frac{20}{10} = 2\]

\[Ответ:x = - 0,2\ \ \ и\ \ x = 2.\]

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

\[Пусть\ \ t = x^{2};\ \ t \geq 0:\]

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

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

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

\[x^{2} = 1\]

\[x = \pm 1\]

\[Ответ:\ \ x = \pm 1.\]

\[\frac{2x^{2} + 11x - 21}{4x^{2} - 9} = \frac{(2x - 3)(x + 7)}{(2x - 3)(2x + 3)} =\]

\[= \frac{x + 7}{2x + 3}\]

\[2x^{2} + 11x - 21 = 2 \cdot (x - 1,5)(x + 7) =\]

\[= (2x - 3)(x + 7)\]

\[D = b^{2} - 4ac = 121 - 4 \cdot 2 \cdot ( - 21) =\]

\[= 121 + 168 = 289\]

\[x_{1} = \frac{- 11 + 17}{4} = \frac{6}{4} = 1,5\]

\[x_{2} = \frac{- 11 - 17}{4} = - \frac{28}{4} = - 7.\]

\[x^{2} + 11x + c = 0\ \ \ \ и\ \ \ \ x_{1} = - 3\]

\[x_{1} + x_{2} = - 11 \Longrightarrow - 3 + x_{2} = - 11 \Longrightarrow\]

\[x_{2} = - 8.\]

\[x_{1} \cdot x_{2} = c \Longrightarrow c = - 3 \cdot ( - 8) \Longrightarrow c = 24.\]

\[Ответ:\ \ x_{2} = - 8\ \ \ и\ \ c = 24.\]

\[\frac{9}{x - 2} - \frac{5}{x} = 2;\ \ \ \ \ \ x \neq 2;\ \ \ x \neq 0\]

\[\frac{9x - 5 \cdot (x - 2)}{x(x - 2)} = 2\]

\[9x - 5x + 10 = 2x(x - 2)\]

\[4x + 10 = 2x^{2} - 4x\]

\[2x^{2} - 8x - 10 = 0\ \ \ \ |\ :2\]

\[x^{2} - 4x - 5 = 0\]

\[x_{1} + x_{2} = 4\]

\[x_{1} \cdot x_{2} = - 5 \Longrightarrow x_{1} = 5\ \ \ и\ \ x_{2} = - 1.\]

\[Ответ:\ \ x = 5\ \ \ и\ x = - 1.\]

\[\left\{ \begin{matrix} x - 2y = 4 \\ \text{xy} = 6\ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \left\{ \begin{matrix} x = 4 + 2y\text{\ \ \ \ \ } \\ (4 + 2y)y = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 4 + 2y\ \ \ \ \ \ \ \ \ \ \ \ \\ 2y^{2} + 4y - 6 = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = 4 + 2y\ \ \ \ \ \ \ \ \\ y^{2} + 2y - 3 = 0 \\ \end{matrix} \right.\ \]

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

\[y_{1} + y_{2} = - 2\]

\[y_{1} \cdot y_{2} = - 3 \Longrightarrow y_{1} = - 3\ \ \ и\ \ \ y_{2} = 1\]

\[\left\{ \begin{matrix} x \neq 4 + 2y \\ \left\lbrack \begin{matrix} y_{1} = - 3 \\ y_{2} = 1\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\lbrack \begin{matrix} \left\{ \begin{matrix} y = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 4 + 2 \cdot ( - 3) \\ \end{matrix}\ \right.\ \\ \left\{ \begin{matrix} y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 4 + 2 \cdot 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \\ \end{matrix} \right.\ \]

\(\left\lbrack \begin{matrix} \left\{ \begin{matrix} y = - 3 \\ x = - 2 \\ \end{matrix}\text{\ \ } \right.\ \\ \left\{ \begin{matrix} y = 1\ \ \ \ \ \\ x = 6\ \ \ \ \\ \end{matrix} \right.\ \\ \end{matrix} \right.\ \)

\[Ответ:\ \ ( - 2;\ - 3);\ \ \ (6;1).\]