Test mathematical formula

[latex]\begin{equation}\label{eq1}r = r_F+ \beta (r_M – r_F) + \epsilon\end{equation}[/latex]
[latex]
\begin{aligned}
\color{red}{\dot{x}} & =\color{blue}{ \sigma(y-x)} \\
\dot{y} & = \rho x – y – xz \\
\dot{z} & = -\beta z + xy
\end{aligned}
[/latex]
[latex]
x^2+y_1+z_{12}^{34} \\
\sin^{-1} (x) \\
\lim\limits_{h->0} \frac{f(x+h)-f(x)}{h} \\
\lim\nolimits_{h\to 0} \frac{f(x+h)-f(x)}{h} \\
f(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \\
\int_0^1 f(x)dx \\
\left[
\begin{array} {lcr}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}
\right]
\\
\left[
\begin{array}{lcr}
a & b \\
c & d
\end{array}
\right] \left(
\begin{array}{lcr}
n \\
k
\end{array}
\right)
[/latex]
[latex]
\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)
\\
\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}
\\
P(E) = {n \choose k} p^k (1-p)^{ n-k}
\\
\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } }
\\
1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|<1$}. \\ \begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned} [/latex]