mirror of
https://github.com/danbulant/notes
synced 2026-06-17 13:31:12 +00:00
Merge branch 'main' of github.com:danbulant/notes
This commit is contained in:
commit
c343336e9e
10 changed files with 506 additions and 153 deletions
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@ -38,16 +310,31 @@
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},
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{
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@ -1,5 +1,18 @@
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{
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"userAdmonitions": {},
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"userAdmonitions": {
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"sentence": {
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"type": "sentence",
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"color": "253, 13, 13",
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"icon": {
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"name": "sticky-note",
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"type": "font-awesome"
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},
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"command": true,
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"title": null,
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"injectColor": true,
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"noTitle": true
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"copyButton": false,
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"version": "6.9.5",
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@ -42,7 +42,7 @@ $C = 355nF = 355 * 10^{-9} F$
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$U = 800V$
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$S = 455cm^2 = 455 * 10^{-2}m^2$
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$C = \frac{Q}{U} = 355 * 10^{-9}F = \frac{Q}{800}$
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$Q = \frac{800 * 355 * 10^{-9}}1 = 2.84 * 10^{-3}C$
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$Q = \frac{800 * 355 * 10^{-9}}1 = 2.84 * 10^{-4}C$
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3. Vypočtěte napětí na kondenzátoru o kapacitě $2.2 nF$, na jehož elektrodách je náboj $7.88 C$.
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@ -14,8 +14,20 @@ Na napětí se používá [Voltmetr](Voltmetr.md).
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### Jak velká je ztráta výkonu v ampérmetru s vnitřním odporem $0,01\Omega$ při měření proudu $110A$?
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$P = U_V * I$
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$R_A = 0.01 \Omega$
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$I = 110A$
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$U_V = R_A * I$
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$P = R_A * I^2$
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### Miliampérmetr s vnitřním odporem $50 \Omega$ má rozsah do $0.01 A$. Vypočítejte velikost bočníku $R_b$ pro rozsah do $50 A$.
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$50 \dots 0.01A$
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$R_b \dots 50A$
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$\frac{50}{0.01} = 5000$
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$R_b = 50 * 5000$
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### Voltmetr s odporem $R_V$ = $400 \Omega$ má rozsah do $10 V$. Jak velký musí být předřadný odpor $R_p$, aby se rozsah voltmetru zvětšil na $250 V$?
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### Proč připojujeme bočník paralelně a předřadný odpor sériově? Vysvětlit můžete i pomocí schématu.
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@ -12,150 +12,6 @@ $$
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$$
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x \neq -1; x \neq 3; x \neq - \frac{1}{2}
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$$
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## Krácení a rozšiřování lomených výrazů
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Lze krátit jen násobení a dělení, ne sčítání a odčítání.
|
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Pro libovolné výrazy $v_1$, $v_2$, $v_3$
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$$
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||||
\frac{v_1 * v_3}{v_2 * v_3} = \frac{v_1}{v_2}
|
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$$
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||||
$$
|
||||
\xRightarrow{krácení}
|
||||
$$
|
||||
$$
|
||||
\xLeftarrow{rozšiřování}
|
||||
$$
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||||
## Sčítání
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||||
- musí se převést na společného jmenovatele
|
||||
$$\frac{7}{5} + \frac{4}{7} = \frac{7}{5}^{*7}_{*7} + \frac{4}{7}^{*5}_{*5*} = \frac{49+20}{35} = \frac{69}{35}$$
|
||||
- musí se stanovit podmínky
|
||||
$$
|
||||
\frac{7}{6} + \frac{1}{x-3} [x\neq3]
|
||||
$$
|
||||
|
||||
```ad-error
|
||||
title: Věta
|
||||
|
||||
Pro libovolné výrazy $V_1;V_2;V_3;V_4$ a pro všechny hodnoty proměnných, pro něž je $V_2 \neq 0; V_4 \neq 0$ platí $\frac{V_1}{V_2} + \frac{V_1}{V_4} = \frac{V_1V_4 + V_3V_2}{V_2V_4}$
|
||||
```
|
||||
---
|
||||
$$
|
||||
\frac{1}{x+1} + \frac{2}{x+3} = \frac{x+3}{(x+1)(x+3)} + \frac{x+1}{(x+1)(x+3)} = \frac{(x+3)+2(x+1)}{(x+1)(x+3)} = \frac{3x+5}{x^2+6x+9}
|
||||
$$
|
||||
---
|
||||
$$
|
||||
[x\neq-1] [x\neq-3]
|
||||
$$
|
||||
$$
|
||||
\frac{3x}{x+2}^{*(x-3)}_{*(x-30)} + \frac{1}{x-3}^{*(x+2)}_{*(x+2)} = \frac{3x^2 - 9x + x + 2}{(x+2)(x-3)} = \frac{3x^2-8x+2}{(x+2)(x-3)}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2][x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{2}{x}+\frac{x}{3} + 4 = \frac{6+x^2+12x}{3x}
|
||||
$$
|
||||
$$
|
||||
[x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{3}{x} + \frac{y}{3x} + \frac{4}{y+1} = \frac{9x(y+1)+y(y+1)+12x}{3x(y+1)} = \frac{(y+1)(9x + y) + 12x}{3x(y+2)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0][y\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x+1}{x} + \frac{x}{x+1} = \frac{(x+1)(x+1)}{x(x+1)} + \frac{x^2}{x(x+1)} = \frac{x^2 + (x+1)(x+1)}{x(x+1)} = \frac{2x^2 + 2x + 1}{x(x+1)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0] [x\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x}{3} - \frac{x}{x+2} - 2 = \frac{x(x+2) - 3x - 3*2(x+2)}{3(x+2)} = \frac{x^2 + 2x - 3x - 6x - 12}{3(x+2)} = \frac{x^2-7x-12}{3x+6}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x+1}{x} + \frac{x-2}{2x} - 2x + 1 = \frac{2(x+1) + x - 2 - 2x2x + 2x}{2x} = \frac{5x-4x^2}{2x} = \frac{x(5-4x)}{2x} = \frac{5-4x}{2} = - 2x + \frac{5}{2}
|
||||
$$
|
||||
$$
|
||||
[x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x}{x+2}-\frac{x+1}{x-3} = \frac{x(x-3) - (x+2)(x+1)}{(x+2)(x-3)} = \frac{x^2-3x-x^2-x-2x-2}{x^2-3x+2x-6} = \frac{-6x-2}{x^2-x-6}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2] [x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{2-3x^2}{x-1} - \frac{2x-1}{2x} - 2 + 3x = \frac{2x(2-3x^2) - (x-1)(2x-1) - 2x(x-1)(3x-2)}{2x(x-1)} = \frac{4x-6x^3 - 2x^2 + x + 1 - 6x^3 + 4x^2 + 6x^2 - 2}{2x(x-1)} = \frac{-12x^2+11x+2}{2x(x-1)}
|
||||
$$
|
||||
$$
|
||||
[x\neq1][x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{y}{y^2 - x^2} - \frac{x}{x-y} = \frac{y}{(y-x)(y+x)} + \frac{x}{y-x} = \frac{y + x(x+y)}{(x+y)(y-x)} = \frac{x^2+xy+y}{y^2-x^2}
|
||||
$$
|
||||
$$
|
||||
[x\neq0][y\neq0][x\neq y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{a+b}{a}-\frac{a}{a-b}+\frac{b^2}{aa-ab} = \frac{(a+b)(a-b) - a^2 + b^2}{a(a-b)} = \frac{a^2-b^2 - a^2 + b^2}{a(a-b)} = 0
|
||||
$$
|
||||
$$
|
||||
[a\neq0][a\neq b]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{8-5x}{8+2x-x^2} - \frac{2x+2}{x^2-3x-4} = \frac{8-5x}{(x+2)(-x+4)} - \frac{2x+2}{(x+1)(x-4)} = \frac{(8-5x)(x+1) + (2x+2)(x+2)}{(x+2)(x+1)(x-4)} = \frac{8x+8-5x^2-5x + 2x^2+4x+4+2x}{(x+2)(x+1)(x-4)} = \frac{-3x^2+9x+12}{(x+2)(x+1)(x-4)}
|
||||
$$
|
||||
$$
|
||||
\frac{3(x+1)(x-4)}{(x+2)(x+1)(x-4)} = \frac{3}{x+2}
|
||||
$$
|
||||
$$
|
||||
[x\neq4][x\neq-2][x\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
8+2x-x^2 = (x+2)(-x+4)
|
||||
$$
|
||||
---
|
||||
### Úkol 17.1.
|
||||
---
|
||||
$$
|
||||
x-y-\frac{x}{x+y} = \frac{(x-y)(x+y)-x}{x+y} = \frac{x^2-y^2-x}{x+y}
|
||||
$$
|
||||
$$
|
||||
[x\neq-y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
1-\frac{4xy}{(x+y)(x+y)} = \frac{(x+y)(x+y)-4xy}{(x+y)(x+y)} = \frac{x^2-2xy+y^2}{(x+y)(x+y)} = \frac{(x-y)(x-y)}{(x+y)(x+y)}
|
||||
$$
|
||||
$$
|
||||
[x\neq-y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x-2}x - \frac{4x+1}{x-3} = \frac{(x-2)(x-3) - x(4x+1)}{x(x-3)} = \frac{-3x(x - 2)}{x(x-3)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0] [x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{3x+8}{x-2} * \frac{x^2 - 4x + 4}{x^2 - 2x + 4} = \frac{(3x+8)(x-2)(x-2)}{(x-2)(x^2 - 2x + 4)} = \frac{3x^3 - 6x^2 - 6x^2 + 12x + 8x^2 - 16x - 16x + 32}{x^3 - 2x^2 + 4x - 2x^2 + 4x + 8} = \frac{3x^3 - 4x^2 - 12x + 32}{x^3 - 4x^2 + 8x + 8} = \frac{x(x(3x-4)-12)+32}{x(x(x-4)+8)+8}
|
||||
$$
|
||||
$$
|
||||
[x\neq2]
|
||||
$$
|
||||

|
||||

|
||||

|
||||
|
|
|
|||
|
|
@ -0,0 +1,32 @@
|
|||
# Dělení a umocňování lomených výrazů
|
||||
```ad-sentence
|
||||
Pro libovolné výrazy $V_1;V_2;V_3;V_4$ a pro všechny hodnoty proměnných, pro něž je $V_2 \neq 0; V_4 \neq 0$, platí: $\frac{V_1}{V_2}*\frac{V_3}{V_4} = \frac{V_1*V_2}{V_3*V_4}$
|
||||
```
|
||||
```ad-sentence
|
||||
Pro libovolné výrazy $V_1;V_2$ a pro libovolné přirozené číslo $k$ a pro všechny hodnoty proměnných, pro něž $V_2 \neq 0$, platí: $(\frac{V_1}{V_2})^k = \frac{V_1^k}{V_2^k}$
|
||||
```
|
||||
```ad-sentence
|
||||
Pro libovolné výrazy $V_1;V_2;V_3;V_4$ a pro všechny hodnoty proměnných, pro něž je $V_2 \neq 0; V_3 \neq 0; V_4 \neq 0$, platí $\frac{V_1}{V_2} / \frac{V_3}{V_4} = \frac{V_1}{V_2} * \frac{V_4}{V_3} = \frac{V_1 * V_4}{V_2 * V_3}$
|
||||
```
|
||||
## Příklady
|
||||
---
|
||||
$$
|
||||
\frac{12a^2b^2}{14x^2y^3}:\frac{18a^2b}{21x^3y^2} = \frac{12a^2b^2 * 21x^3y^2}{14x^2y^3 * 18a^2b} = \frac{bx}{y}
|
||||
$$
|
||||
$$
|
||||
[x\neq0][y\neq0][a\neq0][b\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{(a+b)^2}{a^2-b^2}:\frac{a^2+b^2}{(a-b)^2} = \frac{a+b}{a-b}*\frac{a^2+2ab-b^2}{a^2+b^2} = \frac{(a+b)(a-b)(a-b)}{(a-b)(a^2+b^2)} = \frac{(a+b)(a-b)}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}
|
||||
$$
|
||||
$$
|
||||
[a\neq b][a^2 \neq b^2]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x^3-x^2y}{y+y^2} : \frac{y^3-y^2x}{xy+x} = \frac{x^2(x-y) * x(1+y)}{y(1+y) * y^2(y-x)} = \frac{-x^3}{y^3}
|
||||
$$
|
||||
$$
|
||||
[x \neq 0] [y \neq 0] [y \neq y^2] [xy \neq -x]
|
||||
$$
|
||||
|
|
@ -0,0 +1,13 @@
|
|||
# Krácení a rozšiřování lomených výrazů
|
||||
Lze krátit jen násobení a dělení, ne sčítání a odčítání.
|
||||
|
||||
Pro libovolné výrazy $v_1$, $v_2$, $v_3$
|
||||
$$
|
||||
\frac{v_1 * v_3}{v_2 * v_3} = \frac{v_1}{v_2}
|
||||
$$
|
||||
$$
|
||||
\xRightarrow{krácení}
|
||||
$$
|
||||
$$
|
||||
\xLeftarrow{rozšiřování}
|
||||
$$
|
||||
6
notes/mat/Lomené výrazy/Lomené výrazy.md
Normal file
6
notes/mat/Lomené výrazy/Lomené výrazy.md
Normal file
|
|
@ -0,0 +1,6 @@
|
|||
# Lomené výrazy
|
||||
%% Zoottelkeeper: Beginning of the autogenerated index file list %%
|
||||
- [[mat/Lomené výrazy/Dělení a umocňování lomených výrazů|Dělení a umocňování lomených výrazů]]
|
||||
- [[mat/Lomené výrazy/Krácení a rozšiřování lomených výrazů|Krácení a rozšiřování lomených výrazů]]
|
||||
- [[mat/Lomené výrazy/Sčítání a odčítání|Sčítání a odčítání]]
|
||||
%% Zoottelkeeper: End of the autogenerated index file list %%
|
||||
133
notes/mat/Lomené výrazy/Sčítání a odčítání.md
Normal file
133
notes/mat/Lomené výrazy/Sčítání a odčítání.md
Normal file
|
|
@ -0,0 +1,133 @@
|
|||
# Sčítání a odčítání
|
||||
## Sčítání
|
||||
- musí se převést na společného jmenovatele
|
||||
$$\frac{7}{5} + \frac{4}{7} = \frac{7}{5}^{*7}_{*7} + \frac{4}{7}^{*5}_{*5*} = \frac{49+20}{35} = \frac{69}{35}$$
|
||||
- musí se stanovit podmínky
|
||||
$$
|
||||
\frac{7}{6} + \frac{1}{x-3} [x\neq3]
|
||||
$$
|
||||
|
||||
```ad-sentence
|
||||
Pro libovolné výrazy $V_1;V_2;V_3;V_4$ a pro všechny hodnoty proměnných, pro něž je $V_2 \neq 0; V_4 \neq 0$ platí $\frac{V_1}{V_2} + \frac{V_1}{V_4} = \frac{V_1V_4 + V_3V_2}{V_2V_4}$
|
||||
```
|
||||
---
|
||||
$$
|
||||
\frac{1}{x+1} + \frac{2}{x+3} = \frac{x+3}{(x+1)(x+3)} + \frac{x+1}{(x+1)(x+3)} = \frac{(x+3)+2(x+1)}{(x+1)(x+3)} = \frac{3x+5}{x^2+6x+9}
|
||||
$$
|
||||
---
|
||||
$$
|
||||
[x\neq-1] [x\neq-3]
|
||||
$$
|
||||
$$
|
||||
\frac{3x}{x+2}^{*(x-3)}_{*(x-30)} + \frac{1}{x-3}^{*(x+2)}_{*(x+2)} = \frac{3x^2 - 9x + x + 2}{(x+2)(x-3)} = \frac{3x^2-8x+2}{(x+2)(x-3)}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2][x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{2}{x}+\frac{x}{3} + 4 = \frac{6+x^2+12x}{3x}
|
||||
$$
|
||||
$$
|
||||
[x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{3}{x} + \frac{y}{3x} + \frac{4}{y+1} = \frac{9x(y+1)+y(y+1)+12x}{3x(y+1)} = \frac{(y+1)(9x + y) + 12x}{3x(y+2)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0][y\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x+1}{x} + \frac{x}{x+1} = \frac{(x+1)(x+1)}{x(x+1)} + \frac{x^2}{x(x+1)} = \frac{x^2 + (x+1)(x+1)}{x(x+1)} = \frac{2x^2 + 2x + 1}{x(x+1)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0] [x\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x}{3} - \frac{x}{x+2} - 2 = \frac{x(x+2) - 3x - 3*2(x+2)}{3(x+2)} = \frac{x^2 + 2x - 3x - 6x - 12}{3(x+2)} = \frac{x^2-7x-12}{3x+6}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x+1}{x} + \frac{x-2}{2x} - 2x + 1 = \frac{2(x+1) + x - 2 - 2x2x + 2x}{2x} = \frac{5x-4x^2}{2x} = \frac{x(5-4x)}{2x} = \frac{5-4x}{2} = - 2x + \frac{5}{2}
|
||||
$$
|
||||
$$
|
||||
[x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x}{x+2}-\frac{x+1}{x-3} = \frac{x(x-3) - (x+2)(x+1)}{(x+2)(x-3)} = \frac{x^2-3x-x^2-x-2x-2}{x^2-3x+2x-6} = \frac{-6x-2}{x^2-x-6}
|
||||
$$
|
||||
$$
|
||||
[x\neq-2] [x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{2-3x^2}{x-1} - \frac{2x-1}{2x} - 2 + 3x = \frac{2x(2-3x^2) - (x-1)(2x-1) - 2x(x-1)(3x-2)}{2x(x-1)} = \frac{4x-6x^3 - 2x^2 + x + 1 - 6x^3 + 4x^2 + 6x^2 - 2}{2x(x-1)} = \frac{-12x^2+11x+2}{2x(x-1)}
|
||||
$$
|
||||
$$
|
||||
[x\neq1][x\neq0]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{y}{y^2 - x^2} - \frac{x}{x-y} = \frac{y}{(y-x)(y+x)} + \frac{x}{y-x} = \frac{y + x(x+y)}{(x+y)(y-x)} = \frac{x^2+xy+y}{y^2-x^2}
|
||||
$$
|
||||
$$
|
||||
[x\neq0][y\neq0][x\neq y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{a+b}{a}-\frac{a}{a-b}+\frac{b^2}{aa-ab} = \frac{(a+b)(a-b) - a^2 + b^2}{a(a-b)} = \frac{a^2-b^2 - a^2 + b^2}{a(a-b)} = 0
|
||||
$$
|
||||
$$
|
||||
[a\neq0][a\neq b]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{8-5x}{8+2x-x^2} - \frac{2x+2}{x^2-3x-4} = \frac{8-5x}{(x+2)(-x+4)} - \frac{2x+2}{(x+1)(x-4)} = \frac{(8-5x)(x+1) + (2x+2)(x+2)}{(x+2)(x+1)(x-4)} = \frac{8x+8-5x^2-5x + 2x^2+4x+4+2x}{(x+2)(x+1)(x-4)} = \frac{-3x^2+9x+12}{(x+2)(x+1)(x-4)}
|
||||
$$
|
||||
$$
|
||||
\frac{3(x+1)(x-4)}{(x+2)(x+1)(x-4)} = \frac{3}{x+2}
|
||||
$$
|
||||
$$
|
||||
[x\neq4][x\neq-2][x\neq-1]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
8+2x-x^2 = (x+2)(-x+4)
|
||||
$$
|
||||
---
|
||||
### Úkol 17.1.
|
||||
---
|
||||
$$
|
||||
x-y-\frac{x}{x+y} = \frac{(x-y)(x+y)-x}{x+y} = \frac{x^2-y^2-x}{x+y}
|
||||
$$
|
||||
$$
|
||||
[x\neq-y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
1-\frac{4xy}{(x+y)(x+y)} = \frac{(x+y)(x+y)-4xy}{(x+y)(x+y)} = \frac{x^2-2xy+y^2}{(x+y)(x+y)} = \frac{(x-y)(x-y)}{(x+y)(x+y)}
|
||||
$$
|
||||
$$
|
||||
[x\neq-y]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{x-2}x - \frac{4x+1}{x-3} = \frac{(x-2)(x-3) - x(4x+1)}{x(x-3)} = \frac{-3x(x - 2)}{x(x-3)}
|
||||
$$
|
||||
$$
|
||||
[x\neq0] [x\neq3]
|
||||
$$
|
||||
---
|
||||
$$
|
||||
\frac{3x+8}{x-2} * \frac{x^2 - 4x + 4}{x^2 - 2x + 4} = \frac{(3x+8)(x-2)(x-2)}{(x-2)(x^2 - 2x + 4)} = \frac{3x^3 - 6x^2 - 6x^2 + 12x + 8x^2 - 16x - 16x + 32}{x^3 - 2x^2 + 4x - 2x^2 + 4x + 8} = \frac{3x^3 - 4x^2 - 12x + 32}{x^3 - 4x^2 + 8x + 8} = \frac{x(x(3x-4)-12)+32}{x(x(x-4)+8)+8}
|
||||
$$
|
||||
$$
|
||||
[x\neq2]
|
||||
$$
|
||||
|
|
@ -12,6 +12,7 @@ imagePrefix: 'data/'
|
|||
- [[mat/Druhá odmocnina|Druhá odmocnina]]
|
||||
- [[mat/Funkce/Funkce|Funkce]]
|
||||
- [[mat/Intervaly|Intervaly]]
|
||||
- [[mat/Lomené výrazy/Lomené výrazy|Lomené výrazy]]
|
||||
- [[mat/Lomené výrazy|Lomené výrazy]]
|
||||
- [[mat/Matematika|Matematika]]
|
||||
- [[mat/Mnohočlen|Mnohočlen]]
|
||||
|
|
|
|||
Loading…
Reference in a new issue