ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 42

\[\boxed{\mathbf{42.}}\]

\[\textbf{а)}\ 2\lg{(x - 1)} < \lg{(x + 1)}\]

\[x + 1 > 0\]

\[x > - 1.\]

\[x - 1 > 0\]

\[x > 1.\]

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

\[\lg(x - 1)^{2} < \lg(x + 1)\]

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

\[x^{2} - 2x + 1 < x + 1\]

\[x^{2} - 3x < 0\]

\[x(x - 3) < 0\]

\[0 < x < 3.\]

\[Решение\ неравенства:\]

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

\[Ответ:x \in (1;3).\]

\[\textbf{б)}\ 2\lg{(x + 3)} < \lg{(x + 5)}\]

\[x + 3 > 0\]

\[x > - 3.\]

\[x + 5 > 0\]

\[x > - 5.\]

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

\[\lg(x + 3)^{2} < \lg(x + 5)\]

\[(x + 3)^{2} < x + 5\]

\[x^{2} + 6x + 9 - x - 5 < 0\]

\[x^{2} + 5x + 4 < 0\]

\[x_{1} + x_{2} = - 5;\ \ x_{1} \cdot x_{2} = 4\]

\[x_{1} = - 1;\ \ x_{2} = - 4;\]

\[(x + 4)(x + 1) < 0\]

\[- 4 < x < - 1.\]

\[Решение\ неравенства:\]

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

\[Ответ:x \in ( - 3; - 1).\]