Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 32

\[\boxed{\mathbf{32.}}\]

\[\textbf{а)}\log_{x - 1}\left( x^{2} + 2x \right) =\]

\[= \log_{x - 1}\left( 2x^{2} - 8x + 16 \right)\]

\[\left\{ \begin{matrix} x^{2} + 2x = 2x^{2} - 8x + 16 \\ x^{2} + 2x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - 1 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - 1 \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + 2x = 2x^{2} - 8x + 16\]

\[x^{2} - 10x + 16 = 0\]

\[D_{1} = 25 - 16 = 9\]

\[x_{1} = 5 + 3 = 8;\]

\[x_{2} = 5 - 3 = 2.\]

\[x^{2} + 2x > 0\]

\[x(x + 2) > 0\]

\[x < - 2;\ \ x > 0.\]

\[\left\{ \begin{matrix} x = 8\ \ \ \ \\ x = 2\ \ \ \ \\ x < - 2 \\ x > 0\ \ \ \ \\ x > 1\ \ \ \\ x \neq 2\ \ \ \\ \end{matrix} \right.\ \]

\[x = 8 > 1 - решение.\]

\[Ответ:x = 8.\]

\[\textbf{б)}\log_{x - 2}\left( 2x^{2} - 9x + 21 \right) =\]

\[= \log_{x - 2}\left( x^{2} + x \right)\]

\[\left\{ \begin{matrix} 2x^{2} - 9x + 21 = x^{2} + x \\ x^{2} + x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - 2 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - 2 \neq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2x^{2} - 9x + 21 = x^{2} + x\]

\[x^{2} - 10x + 21 = 0\]

\[D_{1} = 25 - 21 = 4\]

\[x_{1} = 5 + 2 = 7;\]

\[x_{2} = 5 - 2 = 3.\]

\[x^{2} + x > 0\]

\[x(x + 1) > 0\]

\[x < - 1;\ \ x > 0.\]

\[\left\{ \begin{matrix} x = 7\ \ \ \ \\ x = 3\ \ \ \\ x < - 1 \\ x > 0\ \ \ \\ x > 2\ \ \ \\ x \neq 3\ \ \ \\ \end{matrix} \right.\ \]

\[x = 7 > 2 - решение.\]

\[Ответ:x = 7.\]