Страница 256 номер 979, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Решите неравенство: 1) \(\frac{x^2 - 16}{|x + 1|} \le 0\); 2) \(\frac{x^2 - 5x - 14}{|x - 8|} \ge 0\).
1) \(\frac{x^{2} - 16}{|x + 1|} \leqslant 0\)
\(\left\{ \begin{aligned} & x + 1 > 0 \\ & \dfrac{x^{2} - 16}{x + 1} \leqslant 0 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & x > - 1 \\ & \dfrac{(x - 4)(x + 4)}{x + 1} \leqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x > - 1 \\ & (x - 4)(x + 4) \leqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(y = x^{2} - 16\) - парабола, ветви направлены вверх, нули \(- 4\) и \(4\)
\(\Rightarrow \left\{ \begin{aligned} & x \in ( - 1; + \infty) \\ & x \in [ - 4;\ 4] \end{aligned} \right.\ \Rightarrow x \in ( - 1;\ 4]\)
\(\begin{cases} x + 1 < 0 \\ \frac{x^{2} - 16}{- x - 1} \leqslant 0 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x < - 1 \\ & \dfrac{(x - 4)(x + 4)}{x + 1} \geqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x < - 1 \\ & (x - 4)(x + 4) \leqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x \in ( - \infty; - 1) \\ & x \in [ - 4;\ 4] \end{aligned} \right.\ \Rightarrow x \in [ - 4; - 1)\)
Решения \(x \in [ - 4; - 1) \cup ( - 1;\ 4]\)
2) \(\frac{x^{2} - 5x - 14}{|x - 8|} \geqslant 0\)
\(\left\{ \begin{aligned} & x - 8 > 0 \\ & \dfrac{x^{2} - 5x - 14}{x - 8} \geqslant 0 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & x - 8 > 0 \\ & \dfrac{(x + 2)(x - 7)}{x - 8} \geqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x - 8 > 0 \\ & (x + 2)(x - 7) \geqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(y = x^{2} - 5x - 14\) - парабола, ветви направлены вверх, нули \(- 2\) и \(7\)
\(\Rightarrow \left\{ \begin{aligned} & x \in (8; + \infty) \\ & x \in ( - \infty; - 2] \cup [ 7; + \infty) \end{aligned} \right.\ \Rightarrow x \in (8; + \infty)\)
\(\left\{ \begin{aligned} & x - 8 < 0 \\ & \dfrac{x^{2} - 5x - 14}{- x + 8} \geqslant 0 \end{aligned} \right.\ \Rightarrow \begin{cases} x - 8 < 0 \\ \frac{(x + 2)(x - 7)}{x - 8} \leqslant 0 \end{cases}\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x - 8 < 0 \\ & (x + 2)(x - 7) \geqslant 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x \in ( - \infty;\ 8) \\ & x \in ( - \infty; - 2] \cup [ 7; + \infty) \end{aligned} \right.\ \Rightarrow x \in ( - \infty; - 2] \cup [ 7;\ 8)\)
Решения \(x \in ( - \infty; - 2] \cup [ 7;\ 8) \cup (8; + \infty)\)
Ответ: 1) \(x \in [ - 4; - 1) \cup ( - 1;\ 4]\); 2) \(x \in ( - \infty; - 2] \cup [ 7;\ 8) \cup (8; + \infty)\).
