测试 ¶

\[ x ={-b \pm \sqrt{b^2-4ac}\over 2a} \]

\[ \left\{\begin{matrix} x=a + r\text{cos}\theta \\ y=b + r\text{sin}\theta \end{matrix}\right. \]

\[ \begin{array}{l} a\mathop{{x}}\nolimits^{{2}}+bx+c=0 \\ \Delta =\mathop{{b}}\nolimits^{{2}}-4ac \\ \mathop{{x}}\nolimits_{{1,2}}=\frac{{-b \pm \sqrt{{\mathop{{b}}\nolimits^{{2}}-4ac}}}}{{2a}} \\ \mathop{{x}}\nolimits_{{1}}+\mathop{{x}}\nolimits_{{2}}=-\frac{{b}}{{a}} \\ \mathop{{x}}\nolimits_{{1}}\mathop{{x}}\nolimits_{{2}}=\frac{{c}}{{a}} \end{array} \]

\[ \int \frac{1}{\sqrt{1-x^{2}}}\mathrm{d}x= \arcsin x +C \]

\[ \begin{vmatrix} \mathbf{i}& \mathbf{j}& \mathbf{k} \\ \frac{\partial X}{\partial u}& \frac{\partial Y}{\partial u}& 0 \\ \frac{\partial X}{\partial v}& \frac{\partial Y}{\partial v}& 0 \\ \end{vmatrix} \]

\[ \mathop \Phi \nolimits_e = \oint { \mathord{ \buildrel{ \lower3pt \hbox{$ \scriptscriptstyle \rightharpoonup$}} \over E} \cdot {d \mathord{ \buildrel{ \lower3pt \hbox{$ \scriptscriptstyle \rightharpoonup$}} \over S}} = {1 \over {{\varepsilon _0}}}\sum {q} } \]

\[ %此公式需要在设置中开启 unicode 扩展支持 \begin{array}{l} {\huge \unicode{8751}}_\mathbb{S} \mathbf{E} \cdot\mathrm{d}s= \cfrac{Q}{\varepsilon_0} \\ {\huge \unicode{8751}}_\mathbb{S} \mathbf{B} \cdot\mathrm{d}s= 0 \\ {\huge \oint}_{\mathbb{L}}^{} \mathbf{E} \cdot \mathrm{d}l=-\cfrac{\mathrm{d}\Phi _{\mathbf{B}}}{\mathrm{d}t } \\ {\huge \oint}_{\mathbb{L}}^{} \mathbf{B} \cdot \mathrm{d}l=\mu_0I+ \mu_0 \varepsilon_0\cfrac{\mathrm{d}\Phi _{\mathbf{E}}}{\mathrm{d}t } \end{array} \]

\[ %此公式需要在【设置】中开启 mhchem 扩展支持 具体用法请参考【帮助】2.11.2 \ce{SO4^2- + Ba^2+ -> BaSO4 v} \]