Test of MathJax and LaTeX - $\sqrt{4}=2$

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Test of MathJax and LaTeX - $\sqrt{4}=2$

Сообщение Forum_admin » Январь 28, 2017, 7:29 pm

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MathJax
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$\sqrt{4}=2$
${}_{f\sqrt {{a^2} + {b^2}} f}$
$x = {\sqrt{a^2 + b^2} \over a + b}$
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$\int_{a}^{b} f(x)dx = F(b) - F(a)$

When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are:
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$ \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned} \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) \int_{a}^{b} f(x)dx = F(b) - F(a)$

$$ \mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0\end{vmatrix} \int_{a}^{b} f(x)dx = F(b) - F(a)$$

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LaTeX
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[latex]
[latex]
[latex]
[latex]

$\int_{a}^{b} f(x)dx = F(b) - F(a)$

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$ \alpha \beta \gamma $
$$ \alpha \beta \gamma $$

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Исходный код теста

Сообщение Forum_admin » Февраль 19, 2017, 2:41 pm

[img][attachment=0]Example.gif[/attachment][/img]
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Example.gif
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