Difference between revisions of "Упътване за LaTeX"

From Ilianko
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\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>
 
\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>\frac{\sum_{i=1}^{n}}{\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[x_{2007}]{\frac{1}{\sum_{cyclyc}a^{bc}}} - <math>\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{} </math>

Revision as of 16:33, 8 June 2011

Примерни формули:


\frac{1}{2} -

\frac{1}{1+\frac{1}{1+\frac{1}{2}}} -

\sqrt{2} -

\sqrt[3]{2} -

x\ge 1 -

x\le 1 -

1\le x\le \pi -

x\neq 0 -

x_1\ge x_2\ge x_{2007}\ge x_{2007}^n\ge x_{2008}^{n+1} -

\triangle ABC \approx \triangle A_1B_1C_1 -

\sum_{cyclyc} ab = ab+bc+ca -

\sum_{cyclyc}\frac{a}{bc} = \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab} -

\sum_{i=1}^{\infty}\frac{1}{x_i} = \frac{\pi^2}{10} -

\frac{\sum_{i=1}^{n}\sqrt{x_i+\frac{1}{\sqrt{x_i}+1}}}{\sqrt[x_{2007}]{\frac{1}{\sum_{cyclyc}a^{bc}}} -