Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 13

\[\boxed{\mathbf{13.}}\]

\[\textbf{а)}\ |3x + 2| > |2x + 3|\]

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

\[9x^{2} + 12x + 4 > 4x^{2} + 12x + 9\]

\[5x^{2} - 5 > 0\]

\[5(x + 1)(x - 1) > 0\]

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

\[Ответ:x \in ( - \infty; - 1) \cup (1; + \infty).\]

\[\textbf{б)}\ |3x - 4| < |x - 4|\]

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

\[9x^{2} - 24x + 16 < x^{2} - 8x + 16\]

\[8x^{2} - 16x < 0\]

\[8x(x - 2) < 0\]

\[0 < x < 2.\]

\[Ответ:x \in (0;2).\]

\[\textbf{в)}\ \left| \frac{3}{4}x - 2 \right| > \left| 2x - \frac{3}{4} \right|\]

\[\left( \frac{3}{4}x - 2 \right)^{2} > \left( 2x - \frac{3}{4} \right)^{2}\]

\[\frac{9}{16}x^{2} - 3x + 4 > 4x^{2} - 3x + \frac{9}{16}\]

\[3\frac{7}{16}x^{2} - 3\frac{7}{16} < 0\]

\[3\frac{7}{16}\left( x^{2} - 1 \right) < 0\]

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

\[- 1 < x < 1.\]

\[Ответ:x \in ( - 1;1).\]

\[\textbf{г)}\ |5x - 3| > |x - 3|\]

\[(5x - 3)^{2} > (x - 3)^{2}\]

\[25x^{2} - 30x + 9 > x^{2} - 6x + 9\]

\[24x^{2} - 24x > 0\]

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

\[x < 0;\ \ x > 1.\]

\[Ответ:x \in ( - \infty;0) \cup (1; + \infty).\]