Формули MS Word MathType

А це зовсім інший спосіб (не впевнений, що будемо його підтримувати) <wikitex> Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of $Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to the subsets of $Q$ and suppose $f$ satisfies the following conditions:

(i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$,

(ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$,

(iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then

$$ \eqno{(2)}

f(A)+f(B)\ge f(A\cap B) + f(A\cup B).

$$ </wikitex>

[math]\operatorname{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2}\,dt = \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n)!}{n!(2x)^{2n}}[/math]

[math]\sum\nolimits_{1}^{1sdfsd}{sdfg}\frac{-b\pm \sqrt{{{b}^{2}}-4asdfsdf\frac{1}{2}\sqrt{sdf\prod\limits_{234}^{sdf}{sdf}}c}}{2a}[/math]

[math]\frac{n!}{r!\left( n-r \right)!}\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\sqrt{{{b}^{2}}-4ac}\underset{x\to \infty }{\mathop{\lim }}\,\infty \bigcap\limits_{{}}{\Epsilon \sum\nolimits_{\frac{n!}{r!\left( n-r \right)!}}^{{}}{\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}}}[/math]