Λύσεις ανισώσεων

ΛΎΣΕΙΣ 

1.-∣x

${}^2+2x-4|>4\iff$

$$\begin{gathered} {\displaystyle {�}^2+2�-4>4 \\ {�}^2+2�-4<-4} \end{gathered}$$


${\displaystyle {�}^2+2�-4>4}$
${\displaystyle {�}^2+2�-8>0}$
${\displaystyle �=36}$
${\displaystyle {�}_1=2,{�}_2=-4}$
${\displaystyle {�}^2+2�-4>4\iff (�-2)(�+4)>0\iff �\in (-\mathrm{\infty },-4)\cup (2,+\mathrm{\infty })}$


${\displaystyle {�}^2+2�-4<-4}$
${\displaystyle {�}^2+2�<0}$
${\displaystyle �(�+2)<0\iff �\in (-2,0)}$


$$\begin{gathered} {\displaystyle �\in (-\mathrm{\infty },-4)\cup (2,+\mathrm{\infty }) \\ �\in (-2,0) \\ \iff �\in (-\mathrm{\infty },-4)\cup (-2,0)\cup (2,+\mathrm{\infty })} \end{gathered}$$

$2.-$x

${}^2+|6x-24|\le 16$
${\displaystyle \mathrm{\mid }6�-24\mathrm{\mid }\le 16-{�}^2}$

${\displaystyle \overline{\begin{array}{l}6�-24\le 16-{�}^2\\ 6�-24\ge -(16-{�}^2)\end{array}}}$

${\displaystyle \overline{\begin{array}{l}6�-24-16+{�}^2\le 0\\ 6�-24\ge -16+{�}^2\end{array}}}$

${\displaystyle \overline{\begin{array}{l}{�}^2+6�-40\le 0\\ {�}^2-6�+8\le 0\end{array}}}$

${\displaystyle {�}^2+6�-40\le 0}$
${\displaystyle �=196}$
${\displaystyle {�}_1=-10,{�}_2=4}$
${\displaystyle {�}^2+6�-40\le 0\iff (�-4)(�+10)\le 0\iff �\in [-10,4]}$


${\displaystyle {�}^2-6�+8\le 0}$
${\displaystyle �=4}$
${\displaystyle {�}_1=2,{�}_2=4}$
${\displaystyle {�}^2-6�+8\le 0\iff (�-2)(�-4)\le 0\iff �\in [2,4]}$


${\displaystyle \overline{\begin{array}{l}�\in [-10,4]\\ �\in [2,4]\end{array}}\iff �\in [2,4]}$

3.-

$(-\infty ,-3)$	${\displaystyle [-3,4)}$	${\displaystyle (4,+\mathrm{\infty })}$
${\displaystyle �+3}$	-	+	+
${\displaystyle �-4}$	-	-	+

1. ${\displaystyle �\in (-\mathrm{\infty },-3)}$

${\displaystyle -(�+3)-(�-4)\le 11}$
${\displaystyle -�-3-�+4\le 11}$
${\displaystyle -2�+1\le 11}$
${\displaystyle -2�\le 10}$
${\displaystyle �\ge -5}$, но ${\displaystyle �\in (-\mathrm{\infty },-3)\Rightarrow �\in [-5,-3)}$


2. ${\displaystyle �\in [-3,4)}$

${\displaystyle (�+3)-(�-4)\le 11}$
${\displaystyle �+3-�+4\le 11}$
${\displaystyle 0�\le 4-\mathrm{\forall }�}$ но ${\displaystyle �\in [-3,4)}$
${\displaystyle \Rightarrow �\in [-3,4)}$


3. ${\displaystyle �\in (4,+\mathrm{\infty })}$

${\displaystyle (�+3)+(�-4)\le 11}$
${\displaystyle �+3+�-4\le 11}$
${\displaystyle 2�-1\le 11}$
${\displaystyle 2�\le 12}$
${\displaystyle �\le 6}$ ή ${\displaystyle �\in (4,+\mathrm{\infty })\Rightarrow �\in [4,6]}$


 ${\displaystyle �\in [-5,-3)\cup [-3,4)\cup [4,6]\iff �\in [-5,6]}$

4.

$x^2-1=(x-1)(x+1)$

	${\displaystyle (-\mathrm{\infty },-1)}$	${\displaystyle [-1,0)}$	${\displaystyle [0,1)}$	${\displaystyle [1,+\mathrm{\infty })}$
${\displaystyle (�-1)(�+1)}$	+	-	-	+
${\displaystyle �}$	-	-	+	+

1. ${\displaystyle �\in (-\mathrm{\infty },-1)}$
${\displaystyle {�}^2-1<{�}^2-(-�)+1}$
${\displaystyle -1<�+1}$
${\displaystyle �>-2}$, но ${\displaystyle �\in (-\mathrm{\infty },-1)\Rightarrow �\in (-2,-1)}$


 2. ${\displaystyle �\in [-1,0)}$
${\displaystyle -({�}^2-1)<{�}^2-(-�)+1}$
${\displaystyle -{�}^2+1<{�}^2+�+1}$
${\displaystyle -2{�}^2-�<0}$
${\displaystyle 2{�}^2+�>0}$
${\displaystyle �(2�+1)>0\iff �\in (-\mathrm{\infty },-\frac{1}{2})\cup (0,+\mathrm{\infty })}$
ή ${\displaystyle �\in [-1,0)\Rightarrow �\in [-1,-\frac{1}{2})}$


 3. ${\displaystyle �\in [0,1)}$
${\displaystyle -({�}^2-1)<{�}^2-�+1}$
${\displaystyle -{�}^2+1<{�}^2-�+1}$
${\displaystyle 2{�}^2-�>0}$
${\displaystyle �(2�-1)>0\iff �\in (-\mathrm{\infty },0)\cup (\frac{1}{2},+\mathrm{\infty })}$
ή ${\displaystyle �\in [0,1)\Rightarrow �\in [\frac{1}{2},1)}$


 4. ${\displaystyle �\in [1,+\mathrm{\infty })}$
${\displaystyle {�}^2-1<{�}^2-�+1}$
${\displaystyle �<2}$
ή ${\displaystyle �\in [1,+\mathrm{\infty })\Rightarrow �\in [1,2)}$

 ${\displaystyle �\in (-2,-\frac{1}{2})\cup (\frac{1}{2},2)}$