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<math> 2 + 2 = 4 </math>
<math> 2 + 2 = 4 </math>
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
a^2 + b^2 = c^2
e^{i\pi} + 1 = 0
\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}
\frac{d}{dx} \left( x^2 \right) = 2x
(x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k
\lim_{x \to 0} \frac{\sin x}{x} = 1

Revision as of 21:37, 17 October 2024

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[math]\displaystyle{ E=mc^2 }[/math]

[math]\displaystyle{ 2 + 2 = 4 }[/math]

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} a^2 + b^2 = c^2 e^{i\pi} + 1 = 0 \int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \frac{d}{dx} \left( x^2 \right) = 2x (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k \lim_{x \to 0} \frac{\sin x}{x} = 1