一〇二上乳膠測試-2013.09.07

\begin{align*} F(n) &= \sum_{d\mid n}f(d) \\ f(n) &= \sum_{d\mid n}F(d)\mu\left(\frac nd\right) \end{align*}

\begin{align*} z &\longmapsto \sqrt{z\bar z} \end{align*}
 * \bullet|:\mathbb C &\longrightarrow \mathbb R \\

\begin{align*} \nabla\cdot\mathbf E &= \frac\rho{\varepsilon_0} & \int\!\!\!\!\int_{\mathbb{S}}\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\bigcirc\,\,\mathbf E\cdot\mathrm d\mathbf s &= \frac Q{\varepsilon_0} \\ \nabla\cdot\mathbf B &= 0 & \int\!\!\!\!\int_{\mathbb{S}}\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\bigcirc\,\,\mathbf B\cdot\mathrm d\mathbf s &= 0 \\ \nabla\times\mathbf E &= -\frac{\partial\mathbf B}{\partial t} & \oint_{\mathbb L}\mathbf E\cdot\mathrm d\boldsymbol\ell &= -\frac{\mathrm d\Phi_\mathbf B}{\mathrm dt} \\ \nabla\times\mathbf B &= \mu_0\mathbf J+\mu_0\varepsilon_0\frac{\partial\mathbf E}{\partial t} & \oint_{\mathbb L}\mathbf B\cdot\mathrm d\boldsymbol\ell &= \mu_0I+\mu_0\varepsilon_0\frac{\mathrm d\Phi_\mathbf E}{\mathrm dt} \end{align*}

\begin{align*} \|f+g\|_p^p &= \int|f+g|^p\,\mathrm d\mu \\ &\leq \int(|f|+|g|)|f+g|^{p-1}\,\mathrm d\mu \\ &= \int |f||f+g|^{p-1}\,\mathrm d\mu+\int|g||f+g|^{p-1}\,\mathrm d\mu \\ &\leq \left(\left(\int|f|^p\,\mathrm d\mu\right)^{1/p}+\left(\int|g|^p\,\mathrm d\mu\right)^{1/p}\right) \left(\int|f+g|^{(p-1)\left(\frac{p}{p-1}\right)}\,\mathrm d\mu \right)^{1-\frac{1}{p}} \\ &= (\|f\|_p+\|g\|_p)\frac{\|f+g\|_p^p}{\|f+g\|_p} \end{align*}

\begin{align} x &= m^2-n^2 & x' &= d\left(m^2-n^2\right) \\ y &= 2mn & y' &= 2dmn \\ z &= m^2+n^2 & z' &= d\left(m^2+n^2\right) \end{align}

\begin{align} x+y &\geq z \tag{三角不等式}\label{xyz}\\ x^2+y^2 &= z^2 \label{xyz2}\\ x^3+y^3 &= z^3 \tag{費馬大定理}\label{xyz3} \end{align}

\begin{subequations}\begin{align} \frac{\mathrm dx}{\mathrm dt} &= x(\alpha-\beta y) \\ \frac{\mathrm dy}{\mathrm dt} &= -y(\gamma-\delta x) \end{align}\end{subequations}

\begin{subequations} Lotka–Volterra equations \begin{align} \frac{\mathrm dx}{\mathrm dt} &= x(\alpha-\beta y) \\ \frac{\mathrm dy}{\mathrm dt} &= -y(\gamma-\delta x) \end{align} Competitive Lotka–Volterra equations \begin{align} \frac{\mathrm dx}{\mathrm dt} &= rx\left(1-\frac{x+\alpha y}K\right) \\ \frac{\mathrm dy}{\mathrm dt} &= sy\left(1-\frac{y+\beta x}L\right) \end{align} \end{subequations}

\begin{subequations} \begin{align} \intertext{Lotka–Volterra equations} \frac{\mathrm dx}{\mathrm dt} &= x(\alpha-\beta y) \\ \frac{\mathrm dy}{\mathrm dt} &= -y(\gamma-\delta x) \intertext{Competitive Lotka–Volterra equations} \frac{\mathrm dx}{\mathrm dt} &= rx\left(1-\frac{x+\alpha y}K\right) \\ \frac{\mathrm dy}{\mathrm dt} &= sy\left(1-\frac{y+\beta x}L\right) \end{align} \end{subequations}

我們可以透過 \eqref{xyz} 來指稱第一條式子，類似地， \eqref{xyz2} 與 \eqref{xyz3} 將用來指稱第二條與第三條式子.