Ирационални неравенства - задачи с решения

Решение:
[tex]\sqrt{2x - 5} < 3[/tex]

[tex]\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}[/tex] [tex]\Rightarrow x\in [\frac{5}{2}; 7)[/tex]

Отговор: [tex]x\in [\frac{5}{2}; 7)[/tex]

Решение:
[tex]\sqrt{x + 6} < x - 6[/tex]

[tex]\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 [/tex]

[tex]\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}[/tex]

[tex]x^2-13x+30 > 0[/tex]
[tex]x^2-13x+30 = 0[/tex]
[tex]D = 49[/tex]
[tex]x_{1} = 10[/tex] ; [tex]x_{2} = 3[/tex]

[tex]x^2-13x+30 > 0 \Leftrightarrow x\in (-\infty,3) \cup (10,+\infty)[/tex]

[tex]\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)[/tex]

Отговор: [tex]x\in (10,+\infty)[/tex]

Решение:
[tex]\sqrt{3x-7} > 3[/tex]

[tex]\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}[/tex]

Отговор: [tex]x \in \left(\frac{16}{3},+\infin\right)[/tex]

Решение:
[tex]\sqrt{x^2+5x+4} < x+2[/tex]

Нека да решим
[tex]x^2+5x+4 \ge 0[/tex]

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

Ако
[tex]x+2<0[/tex]
[tex]x \in \varnothing[/tex]

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

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

Отговор: [tex]x \in [-1,0)[/tex]

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

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

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

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

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

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

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

Решение:
[tex]\sqrt{x^2+x-12} < 6-x[/tex]

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

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

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

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

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

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

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

Решение:
[tex]\sqrt{7-3x} < -9[/tex]

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

Отговор : [tex]x \in \empty[/tex]

Решение:
[tex]\sqrt{x+2} > 2x-2[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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.[/tex]

[tex]4x^2-9x+2<0[/tex]
[tex]4x^2-9x+2=0[/tex]
[tex]D = 49[/tex]

[tex]x_{1} = 2[/tex], [tex]x_{2} = \frac{1}{4}[/tex]

[tex]4x^2-9x+2<0 \Leftrightarrow x \in (\frac{1}{4},2)[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

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

Отговор: [tex]x \in [-2,2)[/tex]

Решение:
[tex]\sqrt{x+2} > x[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

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

Отговор: [tex]x \in [-2,2)[/tex]

Решение:
[tex]x-3 < \sqrt{x+27}[/tex]

ДМ: [tex]x+27 \ge 0 \Leftrightarrow x \ge -27[/tex]

[tex]\sqrt{x+27} > x-3[/tex]

Ако
[tex]x-3 < 0 \Leftrightarrow x<3[/tex]
[tex]\forall x[/tex] от ДМ
[tex]\Rightarrow x \in [-27,3)[/tex]

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

[tex]x \ge 3 \Rightarrow x \in [3,9)[/tex]
ДМ: [tex]x \ge -27 \Rightarrow x \in [3,9)[/tex]

Отговор: [tex]x \in [-27,3) \cup [3,9) \Leftrightarrow x \in [-27,9)[/tex]

Решение:
[tex]\sqrt{4x-8} > -5[/tex] - това е винаги вярно

[tex]4x-8 \ge 0[/tex]
[tex]4x \ge 8[/tex]
[tex]x \ge 2[/tex]

Отговор: [tex]x \ge 2[/tex]

Решение:
[tex]\sqrt{-x^2+6x-5} > 8-2x[/tex]

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

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

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

Отговор: [tex]x \in (4;5] \cup (3;4] \Leftrightarrow x \in (3;5][/tex]

Решение:
[tex]\sqrt{2x+1} + \sqrt{x+1} < 1[/tex]

ДМ:

[tex]\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)[/tex]

[tex]\sqrt{2x+1} + \sqrt{x+1} < 1[/tex]
[tex]\sqrt{2x+1} < 1 - \sqrt{x+1} \uparrow^2[/tex]
[tex]2x+1 < 1 - 2\sqrt{x+1}+x+1[/tex]
[tex]2\sqrt{x+1} < 1-x[/tex]
[tex]\Leftrightarrow[/tex]
[tex]\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}[/tex]

[tex]x^2-6x-3 > 0[/tex]
[tex]x^2-6x-3 = 0[/tex]
[tex]D = 48[/tex]

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

[tex]\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})[/tex]

но ДМ:

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

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

Решение:
[tex]\sqrt{5x-4} + \sqrt{3x+1} < 3[/tex]

ДМ:

[tex]\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)[/tex]

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

[tex]\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}[/tex]

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

[tex]\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) [/tex]

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

Отговор: [tex]x\in [\frac{4}{5},1)[/tex]

Решение:
[tex]\sqrt{1+x} + \sqrt{1-x} > 1[/tex]

ОДЗ:

[tex]\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][/tex]

И двете страни на неравенството са неотрицателни, затова можем да го повдигнем на квадрат:

[tex]\left(\sqrt{1+x} + \sqrt{1-x}\right)^2 = (1+x) + 2\sqrt{(1+x)(1-x)} + (1-x) = 2 + 2\sqrt{1-x^2}[/tex]

За всяко [tex]x \in [-1,1][/tex] е изпълнено [tex]\sqrt{1-x^2} \ge 0[/tex], следователно

[tex]\left(\sqrt{1+x} + \sqrt{1-x}\right)^2 \ge 2 \Rightarrow \sqrt{1+x} + \sqrt{1-x} \ge \sqrt{2} > 1[/tex]

Неравенството е изпълнено за всяко число от ОДЗ.

Отговор: [tex]x \in [-1,1][/tex]

Решение:
[tex]\sqrt{x^2+3x+2} - \sqrt{x^2-x+1} < 1[/tex]

DM:
[tex]\begin{array}{|l} x^2+3x+2 \ge 0 \\ x^2-x+1 \ge 0 \end{array}[/tex]

[tex]x^2+3x+2 \ge 0[/tex]
[tex]x^2+3x+2 = 0[/tex]

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

[tex]x^2-x+1 \ge 0[/tex]
[tex]x^2-x+1 = 0[/tex]
[tex]D=-3 \Rightarrow[/tex]
[tex]x^2-x+1 \ge 0[/tex] for [tex]\forall x[/tex]

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

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

[tex]\Leftrightarrow[/tex] [tex]\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[/tex] [tex]\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[/tex]

[tex]3x^2+x-1 < 0[/tex]
[tex]3x^2+x-1 = 0[/tex]
[tex]D=13[/tex]

[tex]x_{1/2}=\frac{-1 \pm \sqrt{13}}{6} \approx[/tex]

[tex]x_{1}=\frac{-1 + \sqrt{13}}{6} \approx 0{,}43[/tex]

[tex]x_{2}=\frac{-1 - \sqrt{13}}{6} \approx -0{,}77[/tex]

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

[tex]\Rightarrow[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]x \in (-\infty,\frac{-1 + \sqrt{13}}{6})[/tex]

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

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

Решение:
[tex]\sqrt{2(x+24)} - \sqrt{x-7} \ge \sqrt{x+7}[/tex]

ДМ:
[tex]\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[/tex]

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

[tex]\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[/tex]

[tex]\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}[/tex]

[tex]\Leftrightarrow[/tex] [tex]\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][/tex]

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

Отговор: [tex]x \in [7,25][/tex]

Решение:
[tex](x+2)\sqrt{x^2-2x-3} \ge 0[/tex]

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

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

[tex]\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[/tex], но
ДМ: [tex]x \in (-\infty,-1] \cup [3,+\infty) \Rightarrow x \in [-2,-1] \cup [3,+\infty)[/tex]

[tex]\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 \{-1, 3\} \end{array} \Rightarrow x \in \varnothing[/tex]

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

Решение:
[tex]\frac{\sqrt{5-4x}+7x-6}{x} \ge 2 [/tex]

ДМ:

[tex]\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}[/tex]

[tex]\frac{\sqrt{5-4x}+7x-6}{x} - 2 \ge 0 [/tex]

[tex]\frac{\sqrt{5-4x}+7x-6-2x}{x} \ge 0 [/tex]

[tex]\frac{\sqrt{5-4x}+5x-6}{x} \ge 0 \Leftrightarrow[/tex]

[tex]\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[/tex] [tex]\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.[/tex]

[tex]\sqrt{5-4x} \ge 6-5x[/tex]
[tex]\Leftrightarrow[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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}][/tex]


[tex]\sqrt{5-4x} \le 6-5x \Leftrightarrow[/tex]


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

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

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

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

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

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

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

[tex]\Leftrightarrow[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

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

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

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

[tex]x^2-4x+3 = 0[/tex]
D = 4
[tex]x_{1} = 1[/tex] ; [tex]x_{2} = 3[/tex]
[tex]x^2-4x+3 = (x-1)(x-3)[/tex]

[tex]x^2-3x+2 = 0[/tex]
D = 1
[tex]x_{1} = 2[/tex] ; [tex]x_{2} = 1[/tex]
[tex]x^2-3x+2 = (x-2)(x-1)[/tex]

[tex]\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)[/tex]

[tex]\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)[/tex]

[tex]\sqrt{\frac{x^2-4x+3}{x^2-3x+2}} \le 1 \uparrow^2 [/tex]

[tex]\frac{x^2-4x+3}{x^2-3x+2} \le 1 [/tex]

[tex]\frac{(x-1)(x-3)}{(x-2)(x-1)} \le 1 [/tex]

[tex]\frac{(x-3)}{(x-2)} \le 1 [/tex]

[tex]\frac{(x-3)}{(x-2)}-1 \le 0 [/tex]

[tex]\frac{x-3-(x-2)}{(x-2)} \le 0 [/tex]

[tex]\frac{x-3-x+2}{(x-2)} \le 0 [/tex]

[tex]\frac{-1}{(x-2)} \le 0[/tex] / . -1

[tex]\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)[/tex]

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

Отговор: [tex]x \in [3,+\infty)[/tex]