ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1485

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\[1)\ y = \sqrt{\frac{x - 3}{x + 3}}\]

\[\frac{x - 3}{x + 3} \geq 0\]

\[(x + 3)(x - 3) \geq 0\]

\[x < - 3\ и\ x \geq 3.\]

\[Ответ:\ \ x \in ( - \infty;\ - 3) \cup \lbrack 3;\ + \infty).\]

\[2)\ y = \sqrt{\log_{3}\frac{2x + 1}{x - 6}}\]

\[\frac{2x + 1}{x - 6} > 0\]

\[(2x + 1)(x - 6) > 0\]

\[x < - \frac{1}{2}\ и\ x > 6.\]

\[\log_{3}\frac{2x + 1}{x - 6} \geq 0\]

\[\log_{3}\frac{2x + 1}{x - 6} \geq \log_{3}3^{0}\]

\[\frac{2x + 1}{x - 6} \geq 1\ \ \ \ \ | \bullet (x - 6)^{2}\]

\[(2x + 1)(x - 6) \geq (x - 6)^{2}\]

\[(2x + 1)(x - 6) - (x - 6)^{2} \geq 0\]

\[(x - 6)\left( (2x + 1) - (x - 6) \right) \geq 0\]

\[(x - 6)(2x + 1 - x + 6) \geq 0\]

\[(x + 7)(x - 6) \geq 0\]

\[x \leq - 7\ и\ x > 6.\]

\(Ответ:\ \ x \in ( - \infty;\ - 7\rbrack \cup (6;\ + \infty).\)