notes/notes/mat/Absolutní hodnota.md
2022-03-28 10:10:06 +02:00

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---
tags: [mat]
---
# Absolutní hodnota
$|x|$
```ad-sentence
$|x|$ = $x<0$ ? $-x$ : $x$
```
## Nulový bod
*"bod zlomu"*
$|x|$ -> $x=0$...nulový bod
$|x-3|$ -> x=3...nulový bod
---
$|x-5|=11$ nulový bod $x=5$
$x<5$
$-(x-5)=11$
$-x+5=11 \space | -5$
$-x=6 | *(-1)$
$x=-6$ $[-6\in(-\infty;5)]$
$x \geq 5$
$x-5=11 | -5$
$x=11+5$
$x=16$ $[-16\in<5;+\infty)]$
$K={-6;16}$
zkouška:
$|x-5|=|-6-5|=11$
$|x-5|=|16-5|=11$
---
$|x+3|=63$ nulový bod $x = -3$
$x<3$
$-(x+3)=63$
$-x-3=63$
$-x=66$
$x=66$ $[66\not\in x<3]$
$x \geq 3$
---
$$\textcolor{red}{|x+1|}+\textcolor{green}{|x-1|}=4$$
$\textcolor{red}{nulový \space bod \dots -1}$ $\textcolor{green}{nulový \space bod \dots 1}$
![[data/Absolutní hodnota_2022-02-23 09.05.43.excalidraw.md]]
| - | $x \lt 1$ | $x\ge1$ |
| ----- | --------------- | ------------- |
| $x\lt-1$ | $-(x+1)+-(x-1)=4$ | $\emptyset$ |
| $x\ge-1$ | $(x+1)+-(x-1)=4$ | $(x+1)+(x-1)=4$ |
$k_1$:
$$-(x+1)+-(x-1)=4$$
$$-x-1-x+1=4$$
$$-2x=4$$
$$x=-2$$
$$k_1=-2$$
$k_2: \emptyset$
$k_3$:
$$(x+1)-(x-1)=4$$
$$x+1-x+1=4$$
$$0x+2=4 \space | -2$$
$$0x=2$$
$$k_4 = \emptyset$$
$k_4$:
$$(x+1)+(x-1)=4$$
$$2x=4$$
$$x=2$$
$$k_4 = 2$$
$2 \ge -1 \vee 2\ge 1$
---
$|3x-2|-|2x-3|=3$
nulový bod=$\frac23$ nulový bod=$\frac32$
| - | $x\lt\frac32$ | $x\gt\frac32$ |
| --------------- | --------------------------- | ---------------------- |
| $x \lt \frac23$ | $k_1$: $-(3x-2)- -(2x-3)=3$ | $k_2: \emptyset$ |
| $x\ge\frac23$ | $k_3: (3x-2)- -(2x-3)=3$ | $k_4: (3x-2)-(2x-3)=3$ |
$k_1: -3x+2+2x-3=3$
$-x-1=3$
$-x=4$
$x=-4$
$k_1=\{-4\}$
$-4 < \frac32 \vee x\lt\frac23$
$k_2: \emptyset$
$k_3: 3x-2+2x-3=3$
$5x-5=3$
$5x=8$
$x=\frac85$
$\frac85 \not\in <\frac23;\frac32)$
$k_3 = \emptyset$
---
$3|x-5|+2|x+1|=2$
nulový bod...$5$ nulový bod...$-1$
| - | $x<5$ | $x \ge 5$ |
| -------- | ------------------ | ----------------- |
| $x<-1$ | $-3(x-5)-2(x+1)=2$ | $\emptyset$ |
| $x\ge-1$ | -3(x-5)+2(x+1)=2$ | $3(x-5)+2(x+1)=2$ |
$k_1: -3(x-5)-2(x+1)=2$
$-3x+15-2x-2=2$
$-5x+13=2$
$-5x=-11$
$5x=11$
$x=\frac{11}5$; $\frac{11}5 \not\in (-\infty; -1)$
$k_1=\emptyset$
$k_2: \emptyset$
$k_3: -3(x-5)+2(x+1)=2$
$-3x+15+2x+2=2$
$-x+17=2$
$-x=-15$
$x=15$; $15\not\in <-1; 5)$
$k_3 = \emptyset$
$k_4: 3(x-5)+2(x+1)=2$
$3x-15+2x+2=2$
$5x-13=2$
$5x=15$
$x=3$; $3\not\in <5; \infty)$
$k_4=\emptyset$