Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 35

\[\boxed{\mathbf{35.}}\]

\[(x - 4)(x - 1) > 0\]

\[x < 1;\ \ x > 4.\]

\[M = ( - \infty;1) \cup (4; + \infty).\]

\[3\sqrt{(x - 1)^{2}} - \sqrt{(x - 4)^{2}} < 5\]

\[3|x - 1| - |x - 4| < 5\]

\[x < 1:\]

\[3(1 - x) - (4 - x) < 5\]

\[3 - 3x - 4 + x < 5\]

\[- 2x < 6\]

\[x > - 3.\]

\[x > 4:\]

\[3(x - 1) - (x - 4) < 5\]

\[3x - 3 - x + 4 < 5\]

\[2x < 4\]

\[x < 2.\]

\[Решение\ неравенства:\]

\[x \in ( - 3;1).\]

\[Ответ:\ x \in ( - 3;1).\]

\[(x + 2)(x + 3) > 0\]

\[x < - 3;\ \ x > - 2.\]

\[M = ( - \infty; - 3) \cup ( - 2; + \infty).\]

\[4\sqrt{(x + 2)^{2}} - \sqrt{(x + 3)^{2}} < 9\]

\[4|x + 2| - |x + 3| < 9\]

\[x < - 3:\]

\[- 4(x + 2) - x - 3 < 9\]

\[- 4x - 8 - x - 3 < 9\]

\[- 5x < 20\]

\[x > - 4.\]

\[x > - 2:\]

\[4(x + 2) + x + 3 > 9\]

\[4x + 8 + x + 3 > 9\]

\[5x > - 2\]

\[x > - 0,4.\]

\[Решение\ неравенства:\]

\[x \in ( - \infty; - 4) \cup ( - 0,4; + \infty).\]

\[Ответ:\ x \in ( - \infty; - 4) \cup ( - 0,4; + \infty).\]