Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 17

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\[\textbf{а)}\ \left( x^{2} - 7x + 12 \right)\log_{31}{(x + 5)} = 0\]

\[1)\ \left\{ \begin{matrix} x^{2} - 7x + 12 = 0 \\ x + 5 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 7x + 12 = 0 \\ x > - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - 7x + 12 = 0\]

\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = 4 > - 5 - корень;\ \ \]

\[x_{2} = 3 > - 5 - корень.\]

\[2)\ \left\{ \begin{matrix} \log_{31}{(x + 5)} = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x + 5 = 1\]

\[x = - 4.\]

\[Ответ:x = - 4;x = 3;x = 4.\]

\[\textbf{б)}\ \left( x^{2} + 3x - 4 \right)\log_{32}(3x + 7) = 0\]

\[1)\ \left\{ \begin{matrix} x^{2} + 3x - 4 = 0 \\ 3x + 7 > 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + 3x - 4 = 0 \\ x > - \frac{7}{3} > - 2\frac{1}{3}\ \\ \end{matrix} \right.\ \]

\[x^{2} + 3x - 4 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = - 4 - не\ подходит;\ \ \]

\[x_{2} = 1 > - 2\frac{1}{3} - корень.\]

\[2)\ \left\{ \begin{matrix} \log_{32}{(3x + 7)} = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3x + 7 = 1\]

\[3x = - 6\]

\[x = - 2.\]

\[Ответ:x = - 2;x = 1.\]