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DNS_PROBE_POSSIBLE
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---
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tags: [MOC, cjl, generated, index]
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---
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@ -12,6 +12,7 @@ imagePrefix: 'data/'
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- [[cjl/literatura/Hebrejská literatura/Hebrejská literatura|Hebrejská literatura]]
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- [[cjl/literatura/poznávání žánrů|poznávání žánrů]]
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- [[cjl/literatura/Řecká literatura/Řecká literatura|Řecká literatura]]
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- [[cjl/literatura/slohy/slohy|slohy]]
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- [[cjl/literatura/Slovní spojení|Slovní spojení]]
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- [[cjl/literatura/Tropy a figury|Tropy a figury]]
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- [[cjl/literatura/Základní literaturní pojmy|Základní literaturní pojmy]]
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29
notes/cjl/literatura/slohy/Románské umění.md
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29
notes/cjl/literatura/slohy/Románské umění.md
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# Románské umění
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- na územích bývalé římské říše
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- 1000 - 1250
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## Charakteristika
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- těžkopádné
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- masivní
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- klenba valená nebo křížová
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- okna malá, obloukovitá nebi kulatá (rozeta)
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## Sochařství
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- postavy bez výrazu v obličeji
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- strnulost
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- schematismus
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- později oživení
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## Malířství
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### fresky
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Malby na zdi
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### iluminace
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Kresby v knize, například zdobené první písmeno
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## Hudba
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Trabadúr a žakér mohou být ta samá osoba. Může být více žakérů.
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### Trubadúr
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Ten kdo přednáší
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### Žakér
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Ten kdo doprovází
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## Příklady
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### Rotunda
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(třeba na hoře říp)
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### Bazilika
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Typ kostela
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4
notes/cjl/literatura/slohy/slohy.md
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4
notes/cjl/literatura/slohy/slohy.md
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# slohy
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%% Zoottelkeeper: Beginning of the autogenerated index file list %%
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- [[cjl/literatura/slohy/Románské umění|Románské umění]]
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%% Zoottelkeeper: End of the autogenerated index file list %%
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@ -25,3 +25,107 @@ $$
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$$
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\xLeftarrow{rozšiřování}
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$$
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## Sčítání
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- musí se převést na společného jmenovatele
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$$\frac{7}{5} + \frac{4}{7} = \frac{7}{5}^{*7}_{*7} + \frac{4}{7}^{*5}_{*5*} = \frac{49+20}{35} = \frac{69}{35}$$
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- musí se stanovit podmínky
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$$
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\frac{7}{6} + \frac{1}{x-3} [x\neq3]
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$$
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```ad-error
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title: Věta
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Pro libovolné výrazy $V_1;V_2;V_3;V_4$ a pro všechny hodnoty proměnných, pro něž je $V_2 \neq 0; V_4 \neq 0$ platí $\frac{V_1}{V_2} + \frac{V_1}{V_4} = \frac{V_1V_4 + V_3V_2}{V_2V_4}$
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```
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---
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$$
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\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}
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$$
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---
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$$
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[x\neq-1] [x\neq-3]
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$$
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$$
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\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)}
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$$
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$$
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[x\neq-2][x\neq3]
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$$
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---
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$$
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\frac{2}{x}+\frac{x}{3} + 4 = \frac{6+x^2+12x}{3x}
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$$
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$$
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[x\neq0]
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$$
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---
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$$
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\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)}
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$$
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$$
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[x\neq0][y\neq-1]
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$$
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---
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$$
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\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)}
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$$
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$$
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[x\neq0] [x\neq-1]
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$$
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---
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$$
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\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}
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$$
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$$
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[x\neq-2]
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$$
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---
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$$
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\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}
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$$
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$$
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[x\neq0]
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$$
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---
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$$
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\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}
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$$
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$$
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[x\neq-2] [x\neq3]
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$$
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---
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$$
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\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)}
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$$
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$$
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[x\neq1][x\neq0]
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$$
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---
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$$
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\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}
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$$
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$$
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[x\neq0][y\neq0][x\neq y]
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$$
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---
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$$
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\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
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$$
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$$
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[a\neq0][a\neq b]
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$$
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---
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$$
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\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)}
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$$
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$$
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\frac{3(x+1)(x-4)}{(x+2)(x+1)(x-4)} = \frac{3}{x+2}
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$$
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$$
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[x\neq4][x\neq-2][x\neq-1]
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$$
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---
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$$
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8+2x-x^2 = (x+2)(-x+4)
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$$
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