Страница 119 номер 427, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
1) \(y = \frac{5}{\sqrt{x^{2} - 4x - 12}} + \sqrt{x + 1}\)
\(\begin{cases} x^{2} - 4x - 12 > 0 \\ x + 1 \geqslant 0 \end{cases}\ \Rightarrow \begin{cases} (x + 2)(x - 6) > 0 \\ x \geqslant - 1 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} x \in ( - \infty; - 2) \cup (6; + \infty) \\ x \in [ - 1; + \infty) \end{cases}\ \Rightarrow x \in (6; + \infty)\)
2) \(y = \frac{x - 3}{\sqrt{18 + 3x - x^{2}}} + \frac{8}{x - 5}\)
\(\begin{cases} 18 + 3x - x^{2} > 0 \\ x - 5 \neq 0 \end{cases}\ \Rightarrow \begin{cases} x^{2} - 3x - 18 < 0 \\ x \neq 5 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} (x + 3)(x - 6) < 0 \\ x \neq 5 \end{cases}\ \Rightarrow \begin{cases} x \in ( - 3;\ 6) \\ x \in ( - \infty;\ 5) \cup (5; + \infty) \end{cases}\ \Rightarrow\)
\(\Rightarrow x \in ( - 3;\ 5) \cup (5;\ 6)\)
3) \(y = \sqrt{x^{2} - 5x - 14} - \frac{9}{x^{2} - 81}\)
\(\begin{cases} x^{2} - 5x - 14 \geqslant 0 \\ x^{2} - 81 \neq 0 \end{cases}\ \Rightarrow \begin{cases} (x + 2)(x - 7) \geqslant 0 \\ x \neq \pm 9 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} x \in ( - \infty; - 2] \cup [ 7; + \infty) \\ x \in ( - \infty; - 9) \cup ( - 9;\ 9) \cup (9; + \infty) \end{cases}\)
\(x \in ( - \infty; - 9) \cup ( - 9; - 2] \cup [ 7;\ 9) \cup (9; + \infty)\)
4) \(y = \frac{1}{\sqrt{6 - 7x - 3x^{2}}} + \frac{2}{\sqrt{x + 1}}\)
\(\begin{cases} 6 - 7x - 3x^{2} > 0 \\ x + 1 > 0 \end{cases}\ \Rightarrow \begin{cases} - 3x^{2} - 7x + 6 > 0 \\ x > - 1 \end{cases}\)
\(- 3x^{2} - 7x + 6 > 0\)
\(D = 49 - 4 \cdot 6 \cdot ( - 3) = 49 + 72 = 121\)
\(x_{1} = \dfrac{7 - 11}{- 6} = \dfrac{4}{6} = \dfrac{2}{3};\ x_{2} = \dfrac{7 + 11}{- 6} = - 3\)
\(x \in \left( - 3;\dfrac{2}{3} \right)\)
\(\begin{cases} x \in \left( - 3;\frac{2}{3} \right) \\ x \in ( - 1; + \infty) \end{cases}\ \Rightarrow x \in \left( - 3;\dfrac{2}{3} \right) \cap ( - 1; + \infty) \Rightarrow x \in \left( - 1;\dfrac{2}{3} \right)\)
Ответ: 1) \(x \in (6; + \infty)\); 2) \(x \in ( - 3;\ 5) \cup (5;\ 6)\); 3) \(x \in ( - \infty; - 9) \cup ( - 9; - 2] \cup [ 7;\ 9) \cup (9; + \infty)\); 4) \(x \in \left( - 1;\frac{2}{3} \right)\).
