Ghi ra câu lệnh latex cho ct sau :
$\bot \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} \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}} } } \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} \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) \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x – y – xz \\ \dot{z} & = -\beta z + xy \end{aligned} $
\bot \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} \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}} } } \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} \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) \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x – y – xz \\ \dot{z} & = -\beta z + xy \end{aligned}
\bot \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} \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}} } } \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} \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) \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x – y – xz \\ \dot{z} & = -\beta z + xy \end{aligned} là câu lệnh trong công thức Latex trên.