Ирационални неравенства - задачи с решения

Решение:
$\sqrt{2x - 5} < 3$

$\begin{array}{|l} 2x-5\ge 0 \\ 2x - 5 < 9 \end{array} \Rightarrow \begin{array}{|l} 2x\ge 5 \\ 2x < 14 \end{array} \Rightarrow \begin{array}{|l} x\ge \frac{5}{2} \\ x < 7 \end{array}$ $\Rightarrow x\in [\frac{5}{2}; 7)$

Отговор: $x\in [\frac{5}{2}; 7)$

Решение:
$\sqrt{x + 6} < x - 6$

$\begin{array}{|l} x+6\ge 0 \\ x - 6 > 0 \\ x + 6 < (x - 6)^2 \end{array} \Leftrightarrow \begin{array}{|l} x\ge -6 \\ x > 6 \\ x+6 < x^2-12x+36 \end{array} \Leftrightarrow $

$\begin{array}{|l} x > 6 \\ x+6 < x^2-13x+30 \end{array} \Leftrightarrow \begin{array}{|l} x > 6 \\ x^2-13x+30 > 0 \end{array}$

$x^2-13x+30 > 0$
$x^2-13x+30 = 0$
$D = 49$
$x_{1} = 10$ ; $x_{2} = 3$

$x^2-13x+30 > 0 \Leftrightarrow x\in (-\infty,3) \cup (10,+\infty)$

$\begin{array}{|l} x > 6 \\ x^2-13x+30 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x > 6 \\ x\in (-\infty,3) \cup (10,+\infty) \end{array} \Leftrightarrow x\in (10,+\infty)$

Отговор: $x\in (10,+\infty)$

Решение:
$\sqrt{3x-7} > 3$

$\begin{array}{|l} 3x-7 \ge 0 \\ 3x-7 > 9 \end{array} \Leftrightarrow \begin{array}{|l} 3x \ge 7 \\ 3x > 16 \end{array} \Leftrightarrow \begin{array}{|l} x \ge \frac{7}{3} \\ x > \frac{16}{3} \end{array} \Leftrightarrow x > \frac{16}{3}$

Отговор: $x \in \left(\frac{16}{3},+\infin\right)$

Решение:
$\sqrt{x^2+5x+4} < x+2$

Нека да решим
$x^2+5x+4 \ge 0$

$D=9$
$x_{1}=-4$, $x_{2}=-1$
$x^2+5x+4 \ge 0 \Leftrightarrow (x+1)(x+4) \ge 0 \Leftrightarrow x \in (-\infty,-4)\cup (-1,+\infty)$

Ако
$x+2<0$
$x \in \empty$

Ако
$x+2>0 \Leftrightarrow x>-2$
$\sqrt{x^2+5x+4} < x+2$
$x^2+5x+4 < (x+2)^2$
$x^2+5x+4 < x^2+4x+4$
$x<0$, но
$x>-2 \Rightarrow x \in (-2;0)$

ДМ: $x \in (-\infty,-4) \cup (-1,+\infty) \Rightarrow x \in [-1,0)$

Отговор: $x \in [-1,0)$

Решение:
$\sqrt{x^2-3x-10} < 8-x$
$\begin{array}{|l} x^2-3x-10 \ge 0 \\ 8-x > 0 \\ x^2-3x-10 < (8-x)^2 \end{array}$

$x^2-3x-10 \ge 0$
$x^2-3x-10 = 0$
$D = 49$
$x_{1}=5$, $x_{2}=-2$
$x^2-3x-10 \ge 0 \Leftrightarrow x \in (-\infty,-2] \cup [5,+\infty)$

$\begin{array}{|l} x \in (-\infty,-2] \cup [5,+\infty) \\ x < 8 \\ x^2-3x-10 < 64-16x+x^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-2] \cup [5,8) \\ 13x < 74 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-2] \cup [5,8) \\ x < \frac{74}{13} \approx 5,6 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-2] \cup [5,\frac{74}{13}) \end{array}$

Отговор: $x \in (-\infty,-2] \cup [5,\frac{74}{13})$

Решение:
$\sqrt{x^2+x-12} < 6-x$

$\begin{array}{|l} x^2+x-12 \ge 0 \\ 6-x > 0 \\ x^2+x-12 < (6-x)^2 \end{array}$

$x^2+x-12 \ge 0$
$x^2+x-12 = 0$
$D = 49$
$x_{1}=3$, $x_{2}=-4$
$x^2+x-12 \ge 0 \Leftrightarrow x \in (-\infty,-4] \cup [3,+\infty)$

$\begin{array}{|l} x^2+x-12 \ge 0 \\ 6-x > 0 \\ x^2+x-12 < (6-x)^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-4] \cup [3,+\infty) \\ x < 6 \\ x^2+x-12 < 36-12x+x^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-4] \cup [3,6) \\ 13x < 48 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-4] \cup [3,6) \\ x < \frac{48}{13} \approx 3,7 \end{array}$

$\Leftrightarrow x \in (-\infty,-4] \cup [3,\frac{48}{13})$

Решение:
$\sqrt{7-3x} < -9$

$x \in \empty$
Тъй като:
$\sqrt{7-3x} \ge 0$ за $\forall x$
$-9<0$
$+ < -$ няма решение

Отговор : $x \in \empty$

Решение:
$\sqrt{x+2} > 2x-2$

$\left [ \begin{array}{l} \begin{array}{|l} x+2 \ge 0 \\ 2x-2 < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x+2 \ge 0 \\ 2x-2 \ge 0 \\ x+2 > (2x-2)^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x \ge -2 \\ 2x < 2 /:2 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge -2 \\ 2x \ge 2 /:2 \\ x+2 > 4x^2-8x+4 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x \ge -2 \\ x < 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge -2 \\ x \ge 1 \\ 4x^2-9x+2 < 0 \end{array} \end{array} \right.$

$4x^2-9x+2<0$
$4x^2-9x+2=0$
$D = 49$

$x_{1} = 2$, $x_{2} = \frac{1}{4}$

$4x^2-9x+2<0 \Leftrightarrow x \in (\frac{1}{4},2)$

$\left [ \begin{array}{l} \begin{array}{|l} x \ge -2 \\ x < 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge -2 \\ x \ge 1 \\ 4x^2-9x+2 < 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{} x \in [-2,1) \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge 1 \\ x \in (\frac{1}{4},2) \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} x \in [-2,1) \\ \\ x \in [1,2) \end{array} \right. \Leftrightarrow$ $x \in [-2,1) \cup [1,2) \Leftrightarrow x \in [-2,2)$

Отговор: $x \in [-2,2)$

Решение:
$\sqrt{x+2} > x$

$\left [ \begin{array}{l} \begin{array}{|l} x+2 \ge 0 \\ x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x+2 \ge 0 \\ x \ge 0 \\ x+2 > x^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x \ge -2 \\ x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge -2 \\ x \ge 0 \\ x^2-x-2 < 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in [-2,0) \end{array} \\ \begin{array}{|l} x \ge 0 \\ x \in (-1,2) \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in [-2,0) \end{array} \\ \begin{array}{l} x \in [0,2) \end{array} \end{array} \right. \Leftrightarrow$

$x^2-x-2 < 0$
$x^2-x-2 = 0$
$D = 9$
$x_{1} = 2$ ; $x_{2} = -1$
$x^2-x-2 < 0 \Leftrightarrow x \in (-1,2)$

Отговор: $x \in [-2,2)$

Решение:
$x-3 < \sqrt{x+27}$

ДМ: $x+27 > 0 \Leftrightarrow x > -27$

$\sqrt{x+27} > x-3$

Ако
$x-3 < 0 \Leftrightarrow x<3$
$\forall x$ от ДМ
$\Rightarrow \in (-27,3)$

Ако
$x-3 \ge 0 \Leftrightarrow x \ge 3$
$\sqrt{x+27} > x-3$
$x+27 > (x-3)^2$
$x+27 > x^2-6x+9$
$x^2-7x-18 < 0$
$D = 121$
$x_{1} = -2$ ; $x_{2} = 9$
$x^2-7x-18 < 0 \Leftrightarrow (x+2)(x-9) < 0 \Leftrightarrow x \in (-2,9)$
но

$x \ge 3 \Rightarrow x \in [3,9)$
ДМ: $x > -27 \Rightarrow x \in [3,9)$

Отговор: $x \in (-27,3) \cup [3,9) \Leftrightarrow x \in (-27,9)$

Решение:
$\sqrt{4x-8} > -5$ - това е винаги вярно

$4x-8 \ge 0$
$4x \ge 8$
$x \ge 2$

Отговор: $x \ge 2$

Решение:
$\sqrt{-x^2+6x-5} > 8-2x$

Domain:
$-x^2+6x-5 \ge 0$
$x^2-6x+5 \le 0$
$D = 16$
$x_{1} = 1$ ; $x_{2} = 5$
$x^2-6x+5 \le 0 \Leftrightarrow (x-1)(x-5) \le 0 \Rightarrow x \in [1;5]$

Ако
$8-2x < 0 \Leftrightarrow 8 < 2x \Leftrightarrow x > 4$
$\forall x$ in [1;5]
$\Rightarrow x \in (4;5]$

Ако
$8-2x \ge 0 \Leftrightarrow 8 \ge 2x \Leftrightarrow x \le 4$
$\sqrt{-x^2+6x-5} > 8-2x$
$-x^2+6x-5 > (8-2x)^2$
$-x^2+6x-5 > 64-32x+4x^2$
$5x^2-38x+69 < 0$
$D = 64$
$x_{1} = 4,6$ ; $x_{2} = 3$
$5x^2-38x+69 < 0 \Leftrightarrow 5(x-4,6)(x-3) < 0 \Rightarrow$
$x \in (3;4,6)$

Отговор: $x \in (4;5) \cup (3;4,6) \Leftrightarrow x \in (3;5]$

Решение:
$\sqrt{2x+1} + \sqrt{x+1} < 1$

ДМ:

$\begin{array}{|l} 2x+1 \ge 0 \\ x+1 \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} 2x \ge -1 \\ x \ge -1 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -\frac{1}{2} \\ x \ge -1 \end{array} \Leftrightarrow x \ge -\frac{1}{2} \Leftrightarrow x \in [-\frac{1}{2},+\infty)$

$\sqrt{2x+1} + \sqrt{x+1} < 1$
$\sqrt{2x+1} < 1 - \sqrt{x+1} \uparrow^2$
$2x+1 < 1 - 2\sqrt{x+1}+x+1$
$2\sqrt{x+1} < 1-x$
$\Leftrightarrow$
$\begin{array}{|l} x+1 \ge 0 \\ 1-x \ge 0 \\ 4(x+1) < 1-2x+x^2 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -1 \\ x \le 1 \\ 4x+4 < 1-2x+x^2 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -1 \\ x \le 1 \\ x^2-6x-3 > 0 \end{array}$

$x^2-6x-3 > 0$
$x^2-6x-3 = 0$
$D = 48$

$x_{1/2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm 2\sqrt{3}$
$x_{1} = 3 + 2\sqrt{3} \approx 6,46$
$x_{2} = 3 - 2\sqrt{3} \approx -0,46$
$x^2-6x-3 > 0 \Leftrightarrow x \in (-\infty,3 - 2\sqrt{3}) \cup (3 + 2\sqrt{3},+\infty)$

$\Leftrightarrow \begin{array}{|l} x \ge -1 \\ x \le 1 \\ x^2-6x-3 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -1 \\ x \le 1 \\ x \in (-\infty,3 - 2\sqrt{3}) \cup (3 + 2\sqrt{3},+\infty) \end{array} \Leftrightarrow x \in [-1,3 - 2\sqrt{3})$

но ДМ:

$x \in [-\frac{1}{2},+\infty)$
$\Rightarrow x \in [-\frac{1}{2},3 - 2\sqrt{3})$

Отговор: $x \in [-\frac{1}{2},3 - 2\sqrt{3})$

Решение:
$\sqrt{5x-4} + \sqrt{3x+1} < 3$

ДМ:

$\begin{array}{|l} 5x-4 \ge 0 \\ 3x+1 \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} 5x \ge 4 \\ 3x \ge -1 \end{array} \Leftrightarrow \begin{array}{|l} x \ge \frac{4}{5} \\ x \ge -\frac{1}{3} \end{array} \Leftrightarrow x \ge \frac{4}{5} \Leftrightarrow x \in [\frac{4}{5},+\infty)$

$\sqrt{5x-4} + \sqrt{3x+1} < 3$
$\sqrt{5x-4} < 3 - \sqrt{3x+1} \uparrow^2$
$5x-4 < 9 - 6\sqrt{3x+1} + 3x +1$
$6\sqrt{3x+1} < 14-2x$
$3\sqrt{3x+1} < 7-x$
$\Leftrightarrow$

$\begin{array}{|l} 3x+1 \ge 0 \\ 7-x \ge 0 \\ 9(3x+1) < (7-x)^2 \end{array} \Leftrightarrow \begin{array}{|l} 3x \ge -1 \\ 7 \ge x \\ 27x+9 < 49-14x +x^2 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -\frac{1}{3} \\ x \le 7 \\ x^2-41x+40 > 0 \end{array}$

$x^2-41x+40 > 0$
$x^2-41x+40 = 0$
D = 1521
$x_{1} = 1$ ; $x_{2} = 40$
$x^2-41x+40 > 0 \Leftrightarrow x \in (-\infty,1) \cup (40,+\infty)$

$\begin{array}{|l} x \ge -\frac{1}{3} \\ x \le 7 \\ x^2-41x+40 > 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -\frac{1}{3} \\ x \le 7 \\ x \in (-\infty,1) \cup (40,+\infty) \end{array} \Leftrightarrow x\in [-\frac{1}{3},1) $

но
ДМ:
$x\in [\frac{4}{5},+\infty)$
$\Rightarrow x\in [\frac{4}{5},1)$

Отговор: $x\in [\frac{4}{5},1)$

Решение:
$\sqrt{1+x} + \sqrt{1-x} > 1$

ОДЗ:

$\begin{array}{|l} 1+x \ge 0 \\ 1-x \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -1 \\ x \le 1 \end{array} \Leftrightarrow x \in [-1,1] $

$\sqrt{1+x} > 1 - \sqrt{1-x} $
$1+x > 1-2\sqrt{1-x}+1-x$
$2\sqrt{1-x} > 1-2x$
$\Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 1-2x < 0 \\ 1-x \ge 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} 1-2x \ge 0 \\ 1-x \ge 0 \\ 4(1-x) < (1-2x)^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 2x > 1 \\ x \le 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} 2x \le 1 \\ x \le 1 \\ 4-4x < 1-4x+4x^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x > \frac{1}{2} \\ x \le 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \le \frac{1}{2} \\ x \le 1 \\ 4x^2-3 > 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (\frac{1}{2},1] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \le \frac{1}{2} \\ x \in \left(-\infin, -\frac{\sqrt{3}}{2}\right) \cup \left(\frac{\sqrt{3}}{2}, +\infin\right) \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (\frac{1}{2},1] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x \in \left(-\infin, -\frac{\sqrt{3}}{2}\right) \end{array} \end{array} \right. \Leftrightarrow$

$x \in \left(-\infin, -\frac{\sqrt{3}}{2}\right) \cup (\frac{1}{2},1]$

ДМ: $x \in [-1,1]$

$\Leftrightarrow$
$x \in \left[-1, -\frac{\sqrt{3}}{2}\right) \cup \left(\frac{1}{2},1\right]$

Отговор: $x \in \left[-1, -\frac{\sqrt{3}}{2}\right) \cup \left(\frac{1}{2},1\right]$

Решение:
$\sqrt{x^2+3x+2} - \sqrt{x^2-x+1} < 1$

DM:
$\begin{array}{|l} x^2+3x+2 \ge 0 \\ x^2-x+1 \ge 0 \end{array}$

$x^2+3x+2 \ge 0$
$x^2+3x+2 = 0$

$D=1$
$x_{1}=-1 ; x_{2}=-2$
$x^2+3x+2 \ge 0 \Leftrightarrow x \in (-\infty,-2) \cup (-1,+\infty)$

$x^2-x+1 \ge 0$
$x^2-x+1 = 0$
$D=-3 \Rightarrow$
$x^2-x+1 \ge 0$ for $\forall x$

$\begin{array}{|l} x^2+3x+2 \ge 0 \\ x^2-x+1 \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\ x \in (-\infty,-1) \cup (-2,+\infty) \end{array} \Leftrightarrow x \in (-\infty,-2) \cup (-1,+\infty)$

$\sqrt{x^2+3x+2} - \sqrt{x^2-x+1} < 1$
$\sqrt{x^2+3x+2} < 1 + \sqrt{x^2-x+1} \uparrow^2$
$x^2+3x+2 < 1+2\sqrt{x^2-x+1}+x^2-x+1$
$4x<2\sqrt{x^2-x+1} /:2$
$2x<\sqrt{x^2-x+1}$
$\sqrt{x^2-x+1}>2x$

$\Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{|l} x^2-x+1 \ge 0 \\ 2x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x^2-x+1 \ge 0 \\ 2x \ge 0 \\ x^2-x+1 \ge 4x^2 \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{|l} \forall x \\ x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} \forall x \\ x \ge 0 \\ 3x^2+x-1 < 0 \end{array} \end{array} \right. \Leftrightarrow$

$3x^2+x-1 < 0$
$3x^2+x-1 = 0$
$D=13$

$x_{1/2}=\frac{-1 \pm \sqrt{13}}{6} \approx$

$x_{1}=\frac{-1 + \sqrt{13}}{6} \approx 0,43$

$x_{2}=\frac{-1 - \sqrt{13}}{6} \approx 0,76$

$3x^2+x-1 < 0 \Leftrightarrow x \in (\frac{-1 - \sqrt{13}}{6},\frac{-1 + \sqrt{13}}{6})$

$\Rightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} \forall x \\ x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} \forall x \\ x \ge 0 \\ 3x^2+x-1 < 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (-\infty,0) \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \ge 0 \\ x \in (\frac{-1 - \sqrt{13}}{6},\frac{-1 + \sqrt{13}}{6}) \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (-\infty,0) \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x \in (0,\frac{-1 + \sqrt{13}}{6}) \end{array} \end{array} \right. \Leftrightarrow$

$x \in (-\infty,\frac{-1 + \sqrt{13}}{6})$

Но
ДМ: $x\in (-\infty,-2) \cup (-1,+\infty) \Rightarrow$
$x \in (-\infty,-2) \cup (-1,\frac{-1 + \sqrt{13}}{6})$

Отговор: $x \in (-\infty,-2) \cup (-1,\frac{-1 + \sqrt{13}}{6})$

Решение:
$\sqrt{2(x+24)} - \sqrt{x-7} \ge \sqrt{x+7}$

ДМ:
$\begin{array}{|l} 2(x+24) \ge 0 /: 2 \\ x-7 \ge 0 \\ x+7 \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} x+24 \ge 0 \\ x \ge 7 \\ x \ge -7 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -24 \\ x \ge 7 \\ x \ge -7 \end{array} \Leftrightarrow x \ge 7 \Leftrightarrow x \in [7,+\infty$

$\sqrt{2(x+24)} - \sqrt{x-7} \ge \sqrt{x+7} \uparrow^2$
$(\sqrt{2(x+24)} - \sqrt{x-7})^2 \ge (\sqrt{x+7})^2$
$2(x+24) - 2\sqrt{2(x+24)}\sqrt{x-7} + x-7 \ge x+7$
$2x+48- 2\sqrt{2(x+24)}\sqrt{x-7} -14 \ge 0$
$- 2\sqrt{2(x+24)}\sqrt{x-7} \ge -2x-34$ / . -1
$2\sqrt{2(x+24)}\sqrt{x-7} \le 2x+34$ / : 2
$\sqrt{2(x+24)}\sqrt{x-7} \le x+17$
$\sqrt{2(x+24)(x-7)} \le x+17 \Leftrightarrow$

$\begin{array}{|l} 2(x+24)(x-7) \ge 0 /: 2 \\ x+17 \ge 0 \\ 2(x+24)(x-7) \le x^2+34x+289 \end{array} \Leftrightarrow \begin{array}{|l} (x+24)(x-7) \ge 0 \\ x \ge -17 \\ 2(x^2+24x-7x-168) \le x^2+34x+289 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \in (-\infty,-24] \cup [7,+\infty) \\ x \ge -17 \\ 2(x^2+17x-168) \le x^2+34x+289 \end{array}$

$\Leftrightarrow$ $\begin{array}{|l} x \in [7,+\infty) \\ 2x^2+34x-336 \le x^2+34x+289 \end{array} \Leftrightarrow \begin{array}{|l} x \in [7,+\infty) \\ x^2-625 \le 0 \end{array} \Leftrightarrow \begin{array}{|l} x \in [7,+\infty) \\ (x-25)(x+25) \le 0 \end{array} \Leftrightarrow \begin{array}{|l} x \in [7,+\infty) \\ x \in [-25,25] \end{array} \Leftrightarrow x \in [7,25]$

$x \in [7,25]$, ДМ: $x \in [7,+\infty)$
$\Leftrightarrow x \in [7,25]$

Отговор: $x \in [7,25]$

Решение:
$(x+2)\sqrt{x^2-2x-3} \ge 0$

ДМ:
$x^2-2x-3 \ge 0$
D=16
$x_{1} = -1$ ; $x_{2} = 3$
$x^2-2x-3 \ge 0 \Leftrightarrow (x+1)(x-3) \ge 0 \Leftrightarrow x \in (-\infty,-1] \cup [3,+\infty)$

$(x+2)\sqrt{x^2-2x-3} \ge 0 \Leftrightarrow$
$\begin{array}{|l} (x+2) \ge 0 \\ \sqrt{x^2-2x-3} \ge 0 \end{array}$ or $\begin{array}{|l} (x+2) \le 0 \\ \sqrt{x^2-2x-3} \le 0 \end{array} \Leftrightarrow$

$\begin{array}{|l} (x+2) \ge 0 \\ \sqrt{x^2-2x-3} \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ge -2 \\ \forall x \end{array} \Rightarrow x \ge -2$, но
ДМ: $x \in (-\infty,-1) \cup (3,+\infty) \Rightarrow x \in [-2,-1) \cup (3,+\infty)$

$\begin{array}{|l} (x+2) \le 0 \\ \sqrt{x^2-2x-3} \le 0 \end{array} \Leftrightarrow \begin{array}{|l} x \le -2 \\ x \in \empty \end{array} \Rightarrow x \in \empty$

Отговор: $x \in [-2,-1] \cup [3,+\infty)$

Решение:
$\frac{\sqrt{5-4x}+7x-6}{x} \ge 2 $

ДМ:

$\begin{array}{|l} x \ne 0 \\ 5-4x \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ne 0 \\ 5 \ge 4x \end{array} \Leftrightarrow \begin{array}{|l} x \ne 0 \\ x \le \frac{5}{4} \end{array}$

$\frac{\sqrt{5-4x}+7x-6}{x} - 2 \ge 0 $

$\frac{\sqrt{5-4x}+7x-6-2x}{x} \ge 0 $

$\frac{\sqrt{5-4x}+5x-6}{x} \ge 0 \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} \sqrt{5-4x}+5x-6 \ge 0 \\ x > 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} \sqrt{5-4x}+5x-6 \le 0 \\ x < 0 \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{|l} \sqrt{5-4x} \ge 6-5x \\ x > 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} \sqrt{5-4x} \le 6-5x \\ x < 0 \end{array} \end{array} \right.$

$\sqrt{5-4x} \ge 6-5x$
$\Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 5-4x \ge 0 \\ 6-5x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} 5-4x \ge 0 \\ 6-5x \ge 0 \\ 5-4x \ge (6-5x)^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 5 \ge 4x \\ 6 < 5x \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} 5 \ge 4x \\ 6 \ge 5x \\ 5-4x \ge 36-60x+25x^2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x \le \frac{5}{4} \\ \\ x > \frac{6}{5} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \le \frac{5}{4} \\ \\ x \le \frac{6}{5} \\ \\ 25x^2-56x+31 \le 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (\frac{6}{5},\frac{5}{4}] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \le \frac{6}{5} \\ \\ 25(x-1)(x-\frac{62}{5}) \le 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (\frac{6}{5},\frac{5}{4}] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \le \frac{6}{5} \\ \\ x \in (1,\frac{62}{5}] \end{array} \end{array} \right.\Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in (\frac{6}{5},\frac{5}{4}] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x \in [1,\frac{6}{5}] \end{array} \end{array} \right. \Leftrightarrow x \in [1,\frac{5}{4}]$


$\sqrt{5-4x} \le 6-5x \Leftrightarrow$


$\begin{array}{|l} 5-4x \ge 0 \\ 6-5x > 0 \\ 5-4x \le (6-5x)^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} 5 \ge 4x \\ 6 > 5x \\ 5-4x \le 36-60x+25x^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \le \frac{5}{4} \\ \\ x < \frac{6}{5} \\ \\ 5-4x \le 36-60x+25x^2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \le \frac{6}{5} \\ 25x^2-56x+31 \ge 0 \end{array}\Leftrightarrow$

$\begin{array}{|l} x \le \frac{6}{5} \\ \\ 25x^2-56x+31 \ge 0 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \le \frac{6}{5} \\ \\ 25(x-1)(x-\frac{62}{50}) \ge 0 \end{array} \Leftrightarrow$

$\begin{array}{|l} x \le \frac{6}{5} \\ \\ x \in (-\infty,1) \cup (\frac{62}{50},+\infty) \end{array} \Leftrightarrow$ $x \in (-\infty,1)$

$\Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} \sqrt{5-4x} \ge 6-5x \\ x > 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} \sqrt{5-4x} \le 6-5x \\ x < 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} x \in (1,\frac{5}{4}] \\ x > 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{|l} x \in (-\infty,1) \\ x < 0 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in [1,\frac{5}{4}] \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x \in (-\infty,0) \end{array} \end{array} \right. \Leftrightarrow$

$x \in (-\infty,0) \cup [1,\frac{5}{4}]$

Отговор: $x \in (-\infty,0) \cup [1,\frac{5}{4}]$

Решение:
ДМ:
$\begin{array}{|l} \frac{x^2-4x+3}{x^2-3x+2} \ge 0 \\ \\ x^2-3x+2 \ne 0 \end{array} \Leftrightarrow$

$x^2-4x+3 = 0$
D = 4
$x_{1} = 1$ ; $x_{2} = 3$
$x^2-4x+3 = (x-1)(x-3)$

$x^2-3x+2 = 0$
D = 1
$x_{1} = 2$ ; $x_{2} = 1$
$x^2-3x+2 = (x-2)(x-1)$

$\frac{x^2-4x+3}{x^2-3x+2} \ge 0 \Leftrightarrow \frac{(x-1)(x-3)}{(x-2)(x-1)} \ge 0 \Leftrightarrow \frac{(x-3)}{(x-2)} \ge 0 \Leftrightarrow x \in (-\infty,2) \cup [3,+\infty)$

$\begin{array}{|l} \frac{x^2-4x+3}{x^2-3x+2} \ge 0 \\ \\ x^2-3x+2 \ne 0 \end{array} \Leftrightarrow \begin{array}{|l} x \in (-\infty,2) \cup [3,+\infty) \\ x_{1} \ne 1 ; x_{2} \ne 2 \end{array} \Leftrightarrow x \in (-\infty,1) \cup (1,2) \cup [3,+\infty)$

$\sqrt{\frac{x^2-4x+3}{x^2-3x+2}} \le 1 \uparrow^2 $

$\frac{x^2-4x+3}{x^2-3x+2} \le 1 $

$\frac{(x-1)(x-3)}{(x-2)(x-1)} \le 1 $

$\frac{(x-3)}{(x-2)} \le 1 $

$\frac{(x-3)}{(x-2)}-1 \le 0 $

$\frac{x-3-(x-2)}{(x-2)} \le 0 $

$\frac{x-3-x+2}{(x-2)} \le 0 $

$\frac{-1}{(x-2)} \le 0$ / . -1

$\frac{1}{(x-2)} \ge 0 \Leftrightarrow \begin{array}{|l} x -2 \ge 0 \\ x-2 \ne 0 \end{array} \Leftrightarrow x-2 > 0 \Leftrightarrow x > 2 \Leftrightarrow x \in (2,+\infty)$

Но
ДМ: $x \in (-\infty,1) \cup (1,2) \cup [3,+\infty)$
$\Rightarrow x \in [3,+\infty)$

Отговор: $x \in [3,+\infty)$