Неравенства

[Гост]

| [tex]x^{2 }[/tex] - х - 9| < 3
| [tex]x^{2 }[/tex] - х - 2| [tex]\ge[/tex] 4
[tex]\sqrt{ 5x^{2 } + 4х - 1}[/tex] < 12
[tex]\sqrt{6 - x^{2 } - х}[/tex] [tex]\ge[/tex] 2

[ammornil]

#
[tex]|x^{2}-x-9| < 3 \Rightarrow -3 < x^{2}-x-9 < 3 \Rightarrow \begin{array}{|l} x^{2}-x-9 > -3 \\ x^{2}-x-9 < 3 \end{array}[/tex]

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[tex]\begin{array}{|l} x^{2}-x-6 > 0 \\ x^{2}-x-12 < 0 \end{array}[/tex]
[tex]x^{2}-x-6 \rightarrow D=(-1)^{2}-4\cdot1\cdot(-6)=25 \Rightarrow x_{1,2}=\frac{1\pm 5}{2\cdot 1} \Rightarrow x^{2}-x-6=1\cdot (x-3)\cdot (x+2)[/tex]
[tex]x^{2}-x-12 \rightarrow D=(-1)^{2}-4\cdot1\cdot(-12)=49 \Rightarrow x_{1,2}=\frac{1\pm7}{2\cdot 1} \Rightarrow x^{2}-x-6=1\cdot (x-4)\cdot (x+3)[/tex]

[tex]\begin{array}{|l} x^{2}-x-6 > 0 \\ x^{2}-x-12 < 0 \end{array} \Leftrightarrow \begin{array}{|l} 1\cdot (x-3)\cdot (x+2)> 0 \\ 1\cdot (x-4)\cdot (x+3) < 0 \end{array}[/tex]

Screenshot 2023-02-18 222335.png (8.48 KiB) Прегледано 441 пъти

$$ x \in (-3;-2) \cup (3;4) $$

[ammornil]

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[tex]|x^{2}-x-2| \ge 4 \Leftrightarrow -4 \ge x^{2}-x-2 \cup x^{2}-x-2 \ge 4 \Leftrightarrow x^{2}-x+2 \le 0 \cup x^{2}-x-6 \ge 0[/tex]

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[tex]x^{2}-x+2 \rightarrow D=(-1)^{2}-4\cdot 1\cdot 2=-7 <0 , \begin{cases} a>0 \\ D<0 \end{cases} \Rightarrow x^{2}-x+2 > 0 \hspace{1em} \forall x \in R \Rightarrow x^{2}-x+2 \nleq 0 \Rightarrow x \in \varnothing[/tex]

[tex]x^{2}-x-6 \rightarrow D=(-1)^{2}-4\cdot1 \cdot(-6)=25 \Rightarrow x_{1,2}=\frac{1\pm 5}{2\cdot 1} \Rightarrow x^{2}-x-6=1\cdot (x-3)\cdot (x+2) \ge 4 \Rightarrow x \in (-\infty;-2] \cup [3;+\infty)[/tex]

Screenshot 2023-02-18 224158.png (4.82 KiB) Прегледано 441 пъти

[tex][/tex]
$$ x \in (-\infty;-2] \cup [3;+\infty) $$

[ammornil]

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[tex]\sqrt{5x^{2 }+4x-1}<12[/tex]
[tex]\text{ДМ: } \hspace{2em} 5x^{2 }+4x-1 \ge 0 \rightarrow D=2^{2}-5\cdot (-1)=9 \rightarrow x_{1,2}=\frac{-2\pm 3}{5} \rightarrow \left(x+1\right)\cdot \left(x-\frac{1}{5}\right) \ge 0 \Rightarrow x \in (-\infty;-1] \cup \left[\frac{1}{5};+\infty\right)[/tex]

Screenshot 2023-02-18 225025.png (5.32 KiB) Прегледано 440 пъти

[tex]\sqrt{5x^{2 }+4x-1}<12 \> |(\cdots)^{2} \Leftrightarrow 5x^{2}+4x-1<144 \Leftrightarrow 5x^{2}+4x-145 < 0 \rightarrow D=2^{2}-5\cdot (-145)=729 \rightarrow x_{1,2}=\frac{-2\pm 27}{5} \begin{cases} x_{1}=-\frac{29}{5} \\ x_{2}=5 \end{cases}[/tex]

Screenshot 2023-02-18 225855.png (19.12 KiB) Прегледано 440 пъти

$$ x \in \left(-\frac{29}{5};-1\right] \cup \left[\frac{1}{5};5\right) $$

[ammornil]

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[tex]\sqrt{-x^{2}-x+6}\ge2[/tex]
[tex]\text{ДМ: } \hspace{2em} -x^{2}-x+6 \ge 0 \Leftrightarrow x^{2}+x-6\le 0 \rightarrow D=1^{2}-4\cdot 1\cdot (-6)=25 \rightarrow x_{1,2}=\frac{-1\pm 5}{2\cdot 1} \Rightarrow \left(x+3\right)\cdot (x-2) \le 0 \Rightarrow x \in [-3;2][/tex]

Screenshot 2023-02-18 232411.png (4.44 KiB) Прегледано 439 пъти

[tex]\sqrt{-x^{2}-x+6}\ge2 \>|(\cdots)^{2} \Rightarrow -x^{2}-x+6 \ge 4 \Leftrightarrow x^{2}+x-2 \le 0 \rightarrow D=1^{2}-4\cdot 1\cdot (-2)=9 \rightarrow x_{1,2}=\frac{-1\pm 3}{2\cdot 1} \Rightarrow (x+2)\cdot (x-1) \le 0 \Rightarrow x \in [-2;1][/tex]

Screenshot 2023-02-18 232505.png (10.79 KiB) Прегледано 439 пъти

$$ x \in [-2;1] $$