Упътване за LaTeX - Revision history

Anko at 22:04, 9 December 2012

2012-12-09T22:04:41Z

2012-12-09T22:03:47Z

2012-12-09T22:02:56Z

2012-12-09T22:02:31Z

2012-12-09T22:01:42Z

2012-12-09T22:01:08Z

2012-12-09T21:59:29Z

2012-12-09T21:59:01Z

2011-10-26T15:31:45Z

2011-06-08T16:44:24Z

@@ Line 5: / Line 5: @@
-<math>\frac{1}{2}</math>  \frac{1}{2}
+\frac{1}{2} - <math>\frac{1}{2}</math>
-<math>\frac{1}{1+\frac{1}{1+\frac{1}{2}}}</math> - \frac{1}{1+\frac{1}{1+\frac{1}{2}}}
+\frac{1}{1+\frac{1}{1+\frac{1}{2}}} - <math>\frac{1}{1+\frac{1}{1+\frac{1}{2}}}</math>
-<math>\sqrt{2}</math> \sqrt{2}
+\sqrt{2} - <math>\sqrt{2}</math>

← Older revision		Revision as of 22:03, 9 December 2012
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−	\sum_{cyclyc}ab = ab+bc+ca - <math>\sum_{cyclyc}ab = ab+bc+ca</math>	+	\sum_{cyclyc}ab = ab+bc+ca - <math>\sum_{cyclyc}{ab = ab+bc+ca}</math>

← Older revision		Revision as of 22:02, 9 December 2012
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−	\sum_{cyclyc} ab = ab+bc+ca - <math>\sum_{cyclyc}ab = ab+bc+ca</math>	+	\sum_{cyclyc}ab = ab+bc+ca - <math>\sum_{cyclyc}ab = ab+bc+ca</math>

← Older revision		Revision as of 22:02, 9 December 2012
Line 35:		Line 35:


−	\sum_{cyclyc} ab = ab+bc+ca - <math>\sum_{cyclyc} ab = ab+bc+ca</math>	+	\sum_{cyclyc} ab = ab+bc+ca - <math>\sum_{cyclyc}ab = ab+bc+ca</math>

@@ Line 44: / Line 44: @@
-\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{\sqrt[x_{2007}{\frac{1}{\sum_{cyclyc}a^{bc}}}
+\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{\sqrt[x_{2007}{\frac{1}{\sum_{cyclyc}a^{bc}}} - <math>\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{\sqrt[{2007}]{ \frac{1}{ \sum_{cyclyc}a^{bc}} }} </math>
 ==Връзки:==
 #[http://www.math10.com/forumbg/viewtopic.php?t=3053 Упътване LaTex]
 #http://ilianko.com/files/short-math-guide.pdf

@@ Line 3: / Line 3: @@
 Примерни формули:
 <math></math>
 <math>\frac{1}{2}</math>  \frac{1}{2}
 <math>\frac{1}{1+\frac{1}{1+\frac{1}{2}}}</math> - \frac{1}{1+\frac{1}{1+\frac{1}{2}}}
 <math>\sqrt{2}</math> \sqrt{2}
 \sqrt[3]{2} - <math>\sqrt[3]{2}</math>
 x\ge 1 - <math>x\ge 1</math>
 x\le 1 - <math>x\le 1</math>
 \le x\le \pi - <math>1\le x\le \pi</math>
 x\neq 0 - <math>x\neq 0</math>
 x_1\ge x_2\ge x_{2007}\ge x_{2007}^n\ge x_{2008}^{n+1} - <math>x_1\ge x_2\ge x_{2007}\ge x_{2007}^n\ge x_{2008}^{n+1}</math>
 \triangle ABC \approx \triangle A_1B_1C_1 - <math>\triangle ABC \approx \triangle A_1B_1C_1</math>
 \sum_{cyclyc} ab = ab+bc+ca - <math>\sum_{cyclyc} ab = ab+bc+ca</math>
 \sum_{cyclyc}\frac{a}{bc} = \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab} - <math>\sum_{cyclyc}\frac{a}{bc} = \frac{a}{bc}+\frac{b}{ca}+\sum_{cyclyc}a^{bc}\frac{c}{ab}</math>
 \sum_{i=1}^{\infty}\frac{1}{x_i} = \frac{\pi^2}{10} - <math>\sum_{i=1}^{\infty}\frac{1}{x_i} = \frac{\pi^2}{10} </math>
 \frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{\sqrt[x_{2007}{\frac{1}{\sum_{cyclyc}a^{bc}}}
-- <math>
+- <math>\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}
 {\sqrt[{2007}]{ \frac{1}{ \sum_{cyclyc}a^{bc}} }} </math>

← Older revision		Revision as of 21:59, 9 December 2012
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	{\sqrt[{2007}]{ \frac{1}{ \sum_{cyclyc}a^{bc}} }} </math>		{\sqrt[{2007}]{ \frac{1}{ \sum_{cyclyc}a^{bc}} }} </math>

		+	Връзки:
	[http://www.math10.com/forumbg/viewtopic.php?t=3053 Упътване LaTex]		[http://www.math10.com/forumbg/viewtopic.php?t=3053 Упътване LaTex]
		+	http://ilianko.com/files/short-math-guide.pdf

← Older revision		Revision as of 15:31, 26 October 2011
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		+	Сайтът поддържа Latex синтаксис за въвеждане формули, а иначе използването на Latex, като текстообработваща среда е друг въпрос...
		+
	Примерни формули:		Примерни формули:
	<math></math>		<math></math>