Страница 256 номер 978, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
1) \(\begin{cases} x^{2} - 5x - 6 < 0 \\ x > - 1{,}2 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & (x + 1)(x - 6) < 0 \\ & x \in ( - 1{,}2; + \infty) \end{aligned} \right.\ \Rightarrow \begin{cases} x \in ( - 1;\ 6) \\ x \in ( - 1{,}2; + \infty) \end{cases}\ \Rightarrow\)
\(\Rightarrow x \in ( - 1;\ 6)\)
2) \(\begin{cases} x^{2} - 5x - 6 < 0 \\ x > 0 \end{cases}\ \Rightarrow \begin{cases} x \in ( - 1;\ 6) \\ x \in (0; + \infty) \end{cases}\ \Rightarrow x \in (0;\ 6)\)
3) \(\begin{cases} x^{2} - 5x - 6 < 0 \\ x \geqslant 6 \end{cases}\ \Rightarrow \begin{cases} x \in ( - 1;\ 6) \\ x \in [ 6; + \infty) \end{cases}\ \Rightarrow x \in \varnothing\)
4) \(\begin{cases} x^{2} - 5x - 6 \leqslant 0 \\ x \geqslant 6 \end{cases}\ \Rightarrow \begin{cases} x \in [ - 1;\ 6] \\ x \in [ 6; + \infty) \end{cases}\ \Rightarrow x = 6\)
5) \(\begin{cases} 6x^{2} + x - 2 > 0 \\ x > 1 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & 6\left( x + \frac{2}{3} \right)\left( x - \frac{1}{2} \right) > 0 \\ & x \in (1; + \infty) \end{aligned} \right.\ \Rightarrow\)
\(D = 1 + 4 \cdot 2 \cdot 6 = 49\)
\(x_{1} = \dfrac{- 1 - 7}{12} = - \dfrac{8}{12} = - \dfrac{2}{3}\)
\(x_{2} = \dfrac{- 1 + 7}{12} = \dfrac{6}{12} = \dfrac{1}{2}\)
\(\Rightarrow \left\{ \begin{aligned} & x \in \left( - \infty; - \dfrac{2}{3} \right) \cup \left( \dfrac{1}{2}; + \infty \right) \\ & x \in (1; + \infty) \end{aligned} \right.\ \Rightarrow x \in (1; + \infty)\)
6) \(\begin{cases} 6x^{2} + x - 2 > 0 \\ x > - \frac{1}{3} \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x \in \left( - \infty; - \frac{2}{3} \right) \cup \left( \frac{1}{2}; + \infty \right) \\ & x \in ( - \frac{1}{3}; + \infty) \end{aligned} \right.\ \Rightarrow x \in \left( \frac{1}{2}; + \infty \right)\)
7) \(\begin{cases} 6x^{2} + x - 2 > 0 \\ x > - 2 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x \in \left( - \infty; - \frac{2}{3} \right) \cup \left( \frac{1}{2}; + \infty \right) \\ & x \in ( - 2; + \infty) \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow x \in \left( - 2; - \dfrac{2}{3} \right) \cup \left( \dfrac{1}{2}; + \infty \right)\)
8) \(\begin{cases} 6x^{2} + x - 2 \geqslant 0 \\ x \leqslant \frac{1}{2} \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x \in \left( - \infty; - \frac{2}{3} \right] \cup \left[ \frac{1}{2}; + \infty \right) \\ & x \in \left( - \infty;\frac{1}{2} \right] \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow x \in \left( - \infty; - \dfrac{2}{3} \right] \cup \left\{ \dfrac{1}{2} \right\}\)
Ответ: 1) \(x \in ( - 1;\ 6)\); 2) \(x \in (0;\ 6)\); 3) \(x \in \varnothing\); 4) \(x = 6\); 5) \(x \in (1; + \infty)\); 6) \(x \in \left( \frac{1}{2}; + \infty \right)\); 7) \(x \in \left( - 2; - \frac{2}{3} \right) \cup \left( \frac{1}{2}; + \infty \right)\); 8) \(x \in \left( - \infty; - \frac{2}{3} \right] \cup \left\{ \frac{1}{2} \right\}\).
