Страница 256 номер 980, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(y = \frac{4}{\sqrt{x^{2} + 3x - 10}} + \frac{1}{3x - 9}\)
\(\left\{ \begin{aligned} & x^{2} + 3x - 10 > 0 \\ & 3x - 9 \neq 0 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & (x - 2)(x + 5) > 0 \\ & x \neq 3 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x \in ( - \infty; - 5) \cup (2; + \infty) \\ & x \in ( - \infty;\ 3) \cup (3; + \infty) \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow x \in ( - \infty; - 5) \cup (2;\ 3) \cup (3; + \infty)\)

2) \(y = \frac{6}{\sqrt{12 + x - x^{2}}} - \frac{2}{x^{2} - 4}\)
\(\left\{ \begin{aligned} & 12 + x - x^{2} > 0 \\ & x^{2} - 4 \neq 0 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & x^{2} - x - 12 < 0 \\ & x^{2} \neq 4 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & (x + 3)(x - 4) < 0 \\ & x \in ( - \infty; - 2) \cup ( - 2;\ 2) \cup (2; + \infty) \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \begin{cases} x \in ( - 3;\ 4) \\ x \in ( - \infty; - 2) \cup ( - 2;\ 2) \cup (2; + \infty) \end{cases}\ \Rightarrow\)
\(\Rightarrow x \in ( - 3; - 2) \cup ( - 2;\ 2) \cup (2;\ 4)\)

3) \(y = \sqrt{49 - x^{2}} + \frac{1}{\sqrt{x^{2} + 3x - 4}}\)
\(\left\{ \begin{aligned} & 49 - x^{2} \geqslant 0 \\ & x^{2} + 3x - 4 > 0 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & x^{2} - 49 \leqslant 0 \\ & (x - 1)(x + 4) > 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & x \in [ - 7;\ 7] \\ & x \in ( - \infty; - 4) \cup (1; + \infty) \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow x \in [ - 7; - 4) \cup (1;\ 7]\)

Ответ: 1) \(x \in ( - \infty; - 5) \cup (2;\ 3) \cup (3; + \infty)\); 2) \(x \in ( - 3; - 2) \cup ( - 2;\ 2) \cup (2;\ 4)\); 3) \(x \in [ - 7; - 4) \cup (1;\ 7]\).