## Teste - Să ne obişnuim cu Forumul !

Documente, Tutoriale, Teste de acomodare, Off-Topic
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### Teste - Să ne obişnuim cu Forumul !

Un link rapid, pagină foarte utilă, aici :

http://en.wikipedia.org/wiki/Help:Displaying_a_formula#Larger_expressions

Using $LaTeX$

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### Teste - Să ne obişnuim cu Forumul !

$\sideset{_1^2}{_3^4}\prod_a^b$

${}_1^2\!\Omega_3^4$

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$\sideset{_1^2}{_3^4}\prod_a^b$${}_1^2\!\Omega_3^4$

$\dot{x}, \ddot{x}$

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$\dot{x}, \ddot{x}$

$\hat a ... \bar b ... \vec c ... \widehat{d e f}$

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$\hat a ... \bar b ... \vec c ... \widehat{d e f}$

Arc (workaround) : $\overset{\frown} {AB}$

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$\overset{\frown} {AB}$

$\overbrace{ 1+2+\cdots+100 }^{5050}$ and $\underbrace{ a+b+\cdots+z }_{26}$

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$\overbrace{ 1+2+\cdots+100 }^{5050}$ and $\underbrace{ a+b+\cdots+z }_{26}$

Large (nested) fractions : $\cfrac{2}{c + \cfrac{2}{d + \cfrac{2}{4}}} = a$

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$\cfrac{2}{c + \cfrac{2}{d + \cfrac{2}{4}}} = a$

$\dfrac{2}{c + \dfrac{2}{d + \dfrac{2}{4}}} = a$ (merge foarte bine în orice situaţie cu dfrac ...)

Cancellations in fractions : $\require{cancel}\cfrac{x}{1 + \cfrac{\cancel{y}}{\cancel{y}}} = \cfrac{x}{2}$

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$\require{cancel}\cfrac{x}{1 + \cfrac{\cancel{y}}{\cancel{y}}} = \cfrac{x}{2}$

Bad : $( \dfrac{1}{2} )$ --- Good : $\left ( \dfrac{1}{2} \right )$

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Bad : $( \dfrac{1}{2} )$ --- Good : $\left ( \dfrac{1}{2} \right )$

$\displaystyle \sum_{k=1}^n k^3 = \left [\frac{n(n+1)}{2} \right ]^2$

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$\displaystyle \sum_{k=1}^n k^3 = \left [\frac{n(n+1)}{2} \right ]^2$

Un link rapid, pagină foarte utilă, aici :
http://en.wikipedia.org/wiki/Help:Displaying_a_formula#Larger_expressions

Georgy
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### Teste - Să ne obişnuim cu Forumul !

$a^2+b^2\geqslant 2ab$

carmen.iulia
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### Teste - Să ne obişnuim cu Forumul !

$a^2+b^2+c^2 \geqslant ab+ac+bc$

Marinela
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### Teste - Să ne obişnuim cu Forumul !

$a^3+b^3+c^3+d^3 \geqslant abc+abd+acd+bcd$

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### Teste - Să ne obişnuim cu Forumul !

Georgy » 07 Apr 2015, 11:32 wrote:$a^2+b^2\geqslant 2ab$

Demonstraţia (simplă, cu formula "binom diferenţă la pătrat") foloseşte la următoarea :

carmen.iulia » 07 Apr 2015, 11:34 wrote:$a^2+b^2+c^2 \geqslant ab+ac+bc$

Cât despre aceasta :

Marinela » 07 Apr 2015, 11:35 wrote:$a^3+b^3+c^3+d^3 \geqslant abc+abd+acd+bcd$

... dar mai bine vedeţi generalizarea (pe care am găsit-o acum vreo 5-6 ani) la adresa : Inegalităţi clasice

Condensat, pentru $a_1, a_2, ... , a_n \geqslant 0$ , formula ar arăta aşa :

$\displaystyle \sum_{i=1}^n a_i^{n-1} \geqslant \displaystyle \sum_{i=1}^{n} \displaystyle \prod_{\substack{k=1 \\ k \neq i}}^{n} a_k$

Matt
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### Teste - Să ne obişnuim cu Forumul !

Test - "Registered User" (.. "simple user")

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### Teste - Să ne obişnuim cu Forumul !

Updated Forum Groups ...

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### Teste - Să ne obişnuim cu Forumul !

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