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МатематикаРешебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 48
Задание 48
\[\boxed{\mathbf{48.}}\]
\[x^{2} - 5x - 34 > x + 6 > 0\]
\[\left\{ \begin{matrix} x + 6 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x - 34 > x + 6 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x > - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 6x - 40 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 6x - 40 > 0\]
\[D = 9 + 40 = 49\]
\[x_{1} = 3 + 7 = 10;\]
\[x_{2} = 3 - 7 = - 4.\]
\[(x + 4)(x - 10) > 0\]
\[x < - 4;\ \ x > 10.\]
\[Ответ:x \in ( - 6; - 4) \cup (10; + \infty).\]
\[0 < x^{2} - 16 < 3x + 38\]
\[\left\{ \begin{matrix} x^{2} - 16 > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 16 < 3x + 38 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} (x + 4)(x - 4) > 0 \\ x^{2} - 3x - 54 < 0\ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x < - 4;\ \ \ \ \ \ x > 4 \\ x^{2} - 3x - 54 < 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 3x - 54 < 0\]
\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 54\]
\[x_{1} = 9;\ \ \ x_{2} = - 6.\]
\[(x + 6)(x - 9) < 0\]
\[- 6 < x < 9.\]
\[Ответ:x \in ( - 6;\ - 4) \cup (4;9).\]
\[44 - 2x > x^{2} - 4 > 0\]
\[\left\{ \begin{matrix} x^{2} - 4 > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4 < 44 - 2x \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} (x + 2)(x - 2) > 0 \\ x^{2} + 2x - 48 < 0\ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x < - 2;\ \ \ \ \ \ x > 2 \\ x^{2} + 2x - 48 < 0 \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 48 < 0\]
\[D_{1} = 1 + 48 = 49\]
\[x_{1} = - 1 + 7 = 6;\]
\(x_{2} = - 1 - 7 = - 8.\)
\[(x + 8)(x - 6) < 0\]
\[- 8 < x < 6.\]
\[Ответ:x \in ( - 8; - 2) \cup (2;6).\]
\[0 < 3x + 18 < x^{2} - 22\]
\[\left\{ \begin{matrix} 3x + 18 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 22 > 3x + 18 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x > - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3x - 40 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 3x - 40 > 0\]
\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 40\]
\[x_{1} = 8;\ \ \ x_{2} = - 5.\]
\[(x + 5)(x - 8) > 0\]
\[x < - 5;\ \ \ x > 8.\]
\[Ответ:x \in ( - 6; - 5) \cup (8;\ + \infty).\]
