ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 38

\[\boxed{\mathbf{38.}}\]

\[\textbf{а)}\ \frac{\left| x^{2} - 5x + 4 \right|}{\left| x^{2} - 4 \right|} < 1\]

\[x^{2} - 4 \neq 0\]

\[x^{2} \neq 4\]

\[x \neq \pm 2.\]

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

\[\left| x^{2} - 5x + 4 \right| < \left| x^{2} - 4 \right|\]

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

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

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

\[x^{2} - 5x + 4 = (x - 1)(x - 4).\]

\[(x - 1)^{2}(x - 4)^{2} < \left( x^{2} - 4 \right)^{2}\]

\[- 10x^{3} + 42x^{2} + 40x < 0\]

\[10x^{3} - 42x^{2} - 40x > 0\]

\[x\left( 10x^{2} - 42x - 40 \right) > 0\]

\[10x^{2} - 42x - 40 = 0\ \ \]

\[D_{1} = 441 + 400 = 841\]

\[x_{1} = \frac{21 + 29}{10} = 5;\]

\[x_{2} = \frac{21 - 29}{10} = - \frac{8}{10} = - 0,8;\]

\[(x + 0,8)x(x - 5) > 0\]

\[0 < x < 0,8;\ \ x > 5.\]

\[Решение\ неравенства:\]

\[x \in (0;0,8) \cup (5; + \infty).\]

\[Ответ:x \in (0;0,8) \cup (5; + \infty).\]

\[\textbf{б)}\ \frac{\left| x^{2} - 10x + 9 \right|}{\left| x^{2} - 9 \right|} > 1\]

\[x^{2} - 9 \neq 0\]

\[x^{2} \neq 9\]

\[x \neq \pm 3.\]

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

\[\left| x^{2} - 10x + 9 \right| > \left| x^{2} - 9 \right|\]

\[x^{2} - 10x + 9 = 0\]

\[D_{1} = 25 - 9 = 16\]

\[x_{1} = 5 + 4 = 9;\]

\[x_{2} = 5 - 4 = 1.\]

\[x^{2} - 10x + 9 = (x - 1)(x - 9).\]

\[(x - 1)^{2}(x - 9)^{2} > \left( x^{2} - 9 \right)^{2}\]

\[- 20x^{3} + 136x^{2} - 180x > 0\]

\[20x^{3} - 136x^{2} + 180x < 0\ \ |\ :4\]

\[5x^{3} - 34x^{2} + 45x < 0\]

\[x\left( 5x^{2} - 34x + 45 \right) < 0\]

\[5x^{2} - 34x + 45 = 0\]

\[D_{1} = 289 - 225 = 64\]

\[x_{1} = \frac{17 + 8}{5} = 5;\]

\[x_{2} = \frac{17 - 8}{5} = \frac{9}{5} = 1,8.\]

\[x(x - 1,8)(x - 5) < 0\]

\[x < 0;\ \ 1,8 < x < 5.\]

\[Решение\ неравенства:\]