Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 21
Задание 21
\[\boxed{\mathbf{21.}}\]
\[\textbf{а)}\ \frac{2|x - 3|}{|x - 5|} < \frac{|x - 5|}{|x - 3|}\]
\[|x - 5| \neq 0\]
\[x \neq 5.\]
\[|x - 3| \neq 0\]
\[x \neq 3.\]
\[M = ( - \infty;3) \cup (3;5) \cup (5; + \infty).\]
\[2 \cdot (x - 3)^{2} < (x - 5)^{2}\]
\[2 \cdot \left( x^{2} - 6x + 9 \right) < x^{2} - 10x + 25\]
\[2x^{2} - 12x + 18 - x^{2} + 10x - 25 < 0\]
\[x^{2} - 2x - 7 < 0\]
\[D_{1} = 1 + 7 = 8\]
\[x_{1} = 1 + 2\sqrt{2};\]
\[x_{2} = 1 - 2\sqrt{2}.\]
\[1 - 2\sqrt{2} < x < 1 + 2\sqrt{2}.\]
\[Решение\ неравенства:\]
\[x \in \left( 1 - 2\sqrt{2};3 \right) \cup \left( 3;1 + 2\sqrt{2} \right).\]
\[|x + 1| \neq 0\]
\[x \neq - 1.\]
\[M = ( - \infty; - 1) \cup ( - 1; + \infty).\]
\[2 \cdot (x + 1)^{2} < (x + 3)^{2} - (x - 1)^{2}\]
\[2x^{2} + 4x + 2 < 8x + 8\]
\[2x^{2} - 4x - 6 < 0\ \ \ |\ :2\]
\[x^{2} - 2x - 3 < 0\]
\[D_{1} = 1 + 3 = 4\]
\[x_{1} = 1 + 2 = 3;\]
\[x_{2} = 1 - 2 = - 1;\]
\[(x + 1)(x - 3) < 0\]
\[- 1 < x < 3.\]
\[Решение\ неравенства:\]
\[x \in ( - 1;3).\]
\[Ответ:\ x \in ( - 1;3).\]
\[\textbf{в)}\ \frac{2|x + 2|}{|x + 3|} < \frac{|x + 3|}{|x + 2|}\]
\[|x + 3| \neq 0\]
\[x \neq - 3.\]
\[|x + 2| \neq 0\]
\[x \neq - 2.\]
\[2(x + 2)^{2} < (x + 3)^{2}\]
\[2\left( x^{2} + 4x + 4 \right) < x^{2} + 6x + 9\]
\[2x^{2} + 8x + 8 - x^{2} - 6x - 9 < 0\]
\[x^{2} + 2x - 1 < 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} = - 1 + \sqrt{2};\]
\[x_{2} = - 1 - 2\sqrt{2};\]
\[- 1 - \sqrt{2} < x < - 1 + \sqrt{2}.\]
\[Решение\ неравенства:\]
\[x \in \left( - 1 - \sqrt{2}; - 2 \right) \cup \left( - 2; - 1 + \sqrt{2} \right).\]
\[|x + 5| \neq 0\]
\[x \neq - 5.\]
\[M = ( - \infty; - 5) \cup ( - 5; + \infty).\]
\[2(x + 5)^{2} < (x + 3)^{2} - (x + 7)^{2}\]
\[2x^{2} + 20x + 50 + 8x + 40 < 0\]
\[2x^{2} + 28x + 90 < 0\ \ \ |\ :2\]
\[x^{2} + 14x + 45 < 0\]
\[D_{1} = 49 - 45 = 4\]
\[x_{1} = - 7 + 2 = - 5;\]
\[x_{2} = - 7 - 2 = - 9.\]
\[- 9 < x < - 5.\]
\[Решение\ неравенства:\]
\[x \in ( - 9; - 5).\]
\[Ответ:\ x \in ( - 9; - 5).\]