Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 27

\[\boxed{\mathbf{27.}}\]

\[\textbf{а)}\log_{x}\left( 2x^{2} - 2x - 3 \right) = 2\]

\[\left\{ \begin{matrix} 2x^{2} - 2x - 3 = x^{2} \\ x^{2} > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = 1 + 2 = 3;\]

\[x_{2} = 1 - 2 =\]

\[= - 1 < 0\ (не\ подходит).\]

\[Ответ:x = 3.\]

\[\textbf{б)}\log_{x}\left( x^{3} - 5x + 7 \right) = 3\]

\[\left\{ \begin{matrix} x^{3} - 5x + 7 = x^{3} \\ x^{3} > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- 5x + 7 = 0\]

\[- 5x = - 7\]

\[x = 1,4.\]

\[Ответ:x = 1,4.\]

\[\textbf{в)}\log_{x}{(x + 2)} = 2\]

\[\left\{ \begin{matrix} x + 2 = x^{2} \\ x^{2} > 0\ \ \ \ \ \ \ \\ x > 0\ \ \ \ \ \ \ \ \ \\ x \neq 1\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = 2;\ \ \ \]

\[x_{2} = - 1 < 0\ (не\ подходит).\]

\[Ответ:x = 2.\]

\[\textbf{г)}\log_{x}{(x + 6)} = 2\]

\[\left\{ \begin{matrix} x + 6 = x^{2} \\ x^{2} > 0\ \ \ \ \ \ \ \ \\ x > 0\ \ \ \ \ \ \ \ \ \ \\ x \neq 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3;\]

\[x_{2} = - 2 < 0\ (не\ подходит).\]

\[Ответ:x = 3.\]