LaTeX Help - Tutorial

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LaTeX Help - Tutorial

Postby Admin » 11 Apr 2015, 19:11

$\boxed{a^2 + b^2 \geqslant 2ab}$

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$\boxed{a^2 + b^2 \geqslant 2ab}$


$\displaystyle \left\{ \frac{x}{y} \right\}$

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\displaystyle \left\{ \frac{x}{y} \right\}


Identitatea lui Hermite - scrisă în LaTeX cu paranteze drepte mari - \Big[ ... \Big] - (Variante: big, Big, bigg, Bigg)

$\displaystyle \Big [x\Big] + \Big[x+\frac{1}{n}\Big] + \Big[x+\frac{2}{n}\Big] + ... + \Big[x+\frac{n-1}{n}\Big] = \Big[nx\Big]$

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$\displaystyle ... \Big[x+\frac{n-1}{n}\Big] ...$


(A nu se uita \displaystyle - fiind fracții în interior...)

$|x|=
\left\{
\begin{matrix} x, ~~~ if ~~ x\geqslant 0\\ -x, ~~~ if ~~~ x<0 \end{matrix}
\right.$

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$|x|=
\left\{
  \begin{matrix} x, ~~~ if ~~ x\geqslant 0\\ -x, ~~~ if ~~~ x<0 \end{matrix}
\right.$



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Postby Admin » 27 Jun 2016, 17:43

$\sqrt[p]{a}$ versus $\sqrt[\leftroot{3}\uproot{3}p]{a}$ versus $\sqrt[\leftroot{-1}\uproot{3}p]{a}$

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$\sqrt[p]{a}$ versus $\sqrt[\leftroot{3}\uproot{3}p]{a}$ versus $\sqrt[\leftroot{-1}\uproot{3}p]{a}$

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Postby Admin » 30 Jun 2016, 22:24

\displaystyle în cazul unor radicali:


$\sqrt \frac{n^2+1}{2^n-7}$ ...respectiv $\displaystyle \sqrt \frac{n^2+1}{2^n-7}$

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CODURI FOLOSITE :
- La primul radical : $\sqrt \frac{n^2+1}{2^n-7}$
- La al 2-lea radical : $\displaystyle \sqrt \frac{n^2+1}{2^n-7}$

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Postby Admin » 04 Jul 2016, 16:31

$\bar{\mathbb{R}} ~~~ and ~~~ \overline{\mathbb{R}}$

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$\bar{\mathbb{R}} ~~~ and ~~~ \overline{\mathbb{R}}$


$\overline{a_1a_2...a_n}_{(10)} ~ = ~ a_1\cdot10^{n-1} + a_2\cdot10^{n-2} + ... + a_{n-1}\cdot10^1 + a_n\cdot10^0$     (folosind \overline)

$\bar{a_1a_2...a_n} ~ = ~ ...$     (folosind \bar)

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Postby Admin » 06 Jul 2016, 22:35

media armonică $\leqslant$ media geometrică $\leqslant$ media aritmetică $\leqslant$ media pătratică

$\displaystyle \dfrac{n}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + ... + \dfrac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \dfrac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\dfrac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$

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$\displaystyle \dfrac{n}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + ... + \dfrac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \dfrac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\dfrac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$


... în loc de:

$\frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \frac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\frac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$

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$\frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \frac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\frac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$



Ce apar în plus (la prima) sunt: prefixarea cu \displaystyle și folosirea \dfrac în loc de \frac.
Nu e suficientă prefixarea întregii expresii cu \displaystyle, acesta influențând la «nivelul» la care se află, iar pentru fracțiile de la numitorul mediei armonice trebuie iar \displaystyle (o dată). Dar e simplu cu \dfrac în loc de \frac.

$\displaystyle \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \frac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\frac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$

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$\displaystyle \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}} \leqslant \sqrt[n]{a_1 a_2 ... a_n} \leqslant \frac{a_1 + a_2 + ... + a_n}{n} \leqslant \sqrt{\frac{a_1^2 + a_2^2 + ... + a_n^2}{n}}$

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Postby Admin » 13 Aug 2016, 00:36

Modul «mare» (și în general folosirea/prefixarea cu \left... și \right...)

$\left|\frac{42}{1+x^4}\right|$

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\left|\frac{42}{1+x^4}\right|


Iar cu \displaystyle la început:

$\displaystyle \left|\frac{42}{1+x^4}\right|$

Alte variante reies din folosirea big, Big, bigg și Bigg.



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Postby Admin » 19 Aug 2016, 16:15

$ \Big||a|-|b|\Big| \leqslant |a \pm b| \leqslant |a| + |b|$

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\Big||a|-|b|\Big| \leqslant |a \pm b| \leqslant |a| + |b|


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