LaTex中的组合表达式|博客编辑


前言

理科老师中总会有一部分人,有一肚子的感悟和反思,总想找个合适的工具表达而不得,本篇博文中提到的表达式或许可以参考;

使用方法:将鼠标放置到符号表达式上点击右键,依次点击Show Math As\(\Rightarrow\) Text Commands,然后复制出现的框中的字符即可;

自动编号

\[\begin{equation}\label{*} x^{y^z}=(1+e^x-2xy^w) \end{equation}\]

公式的引用:比如,由上数据公式\((\ref{*})\)可以得到相应的结论;[自动编号]

组合表达

\(\Delta ABC\sim\Delta XYZ\;\;\;\)\(\sqrt{3}\approx1.732050808\ldots\)\(\sqrt[4]{16}=2\)\(\sum\limits_{k=1}^n{x_k}\)\(\prod\limits_{k=1}^n{x_k}\)\(p(x_{S}|x_{S\mkern-8.5mu/})\)\(\stackrel{一一对应}{\Longleftrightarrow}\)

\(y=A\sin(\omega\cdot x+\phi)+k\)\(\sqrt[4]{x}\)\(0.\dot{3}\dot{6}\)\(\sin\!\cfrac{\pi}{3}\) = \(\sin60^\circ\)=\(\cfrac{\sqrt{3}}{2}\)\(0\stackrel{+abc}{\Longrightarrow}1\)\(0\stackrel{a}{\Longleftrightarrow}1\)\(\cfrac{\underline{\;加热\;}}{炮制}\)

\(\cos\langle\overrightarrow{CM}, \overrightarrow{BD}\rangle=\cfrac{\sqrt{3}}{6}\);向量加粗:\(\bm{a}\)\(\textbf{频率}=\cfrac{\textbf{频数}}{\textbf{样本容量}}\)

直线\(l\)的参数方程为\(\left\{\begin{array}{l}{x=2+\cos\alpha\cdot t}\\{y=2+\sin\alpha\cdot t}\end{array}\right.\)\((t为参数)\);直线\(l\)的参数方程为\(\begin{cases}x=2+cos\alpha\cdot t\\y=2+sin\alpha\cdot t\end{cases}(t为参数)\)

\(m\)的取值范围为\(m\in \left\{ m\Big|\cfrac{1}{2}<m<\cfrac{2}{3} \right\}\)\(\xleftarrow{123}\); 极限符号 \(\lim\limits_{x\to 1^+} f(x)=\lim\limits_{x\to 1^+}\cfrac{x}{x^2+3x+1}\)

定积分符号\(\displaystyle\int_{-1}^{2} e^x\, dx=e^x\bigg|_{1}^{2}=e^2-e^{-1}\);$\cfrac{df}{dx}\bigg|_{x = x_0} $

二项分布符号 \(X\sim B\left(3,\cfrac{1}{5}\right)\);不等式组\(\begin{cases} &-2 < m-1 < 2 \\ &-2 <2m-1 <2 \\ &m-1<1-2m\end{cases}\)

\(\begin{gather*}f(x+4) &=-f(x)\\f(-x)&=-f(x)\end{gather*}\) \(\Bigg\}\Longrightarrow f(x+4)=f(-x)\Longrightarrow\)对称轴是\(x=2\)

\(\left.\begin{array}{1}\text{if n is even:}&n/2\\\text{if n is odd:}&3n+1\end{array}\right\}=f(n)\)\(\left\{\begin{array}{ll}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y&=d_2\\a_3x+b_3y+c_3z&=d_3\end{array}\right.\)

\(f(n)=\begin{cases}\frac{n}{2},&\text{if n is even}\\[2ex]3n+1,&\text{if n is odd}\end{cases}\)\(\left\{\begin{array}{ll}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y&=d_2\\a_3x+b_3y+c_3z&=d_3\end{array}\right.\)

\(\begin{cases}-2<m-1<2&666\\x+y=1\\ m-1<1-2m\end{cases}\)\(\left\{\begin{array}{l}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y&=d_2\\a_3x+b_3y+c_3z&=d_3\end{array}\right.\)

\(\left\{\begin{array}{r}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y&=d_2\\a_3x+b_3y+c_3z&=d_3\end{array}\right.\)\(\left\{\begin{aligned}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y&=d_2\\a_3x+b_3y+c_3z&=d_3\end{aligned}\right.\)

\(\bbox[yellow,10px]{e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\qquad(1)}\)\(\bbox[10px,yellow,border:2px dashed red]{e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\qquad(1)}\)\(\bbox[5px,border:2px solid red]{e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\qquad(1)}\)

\(f(x)=\begin{cases}x^2 &x\leq 0\\ 3x+1 &x>0 \end{cases}\);$ |AB|\xlongequal[韦达定理]{直角坐标系下}|x_1-x_2|\sqrt{1+k^2}$

\(f_n(x)=\underbrace{f\{f[f\cdots f}_{n个}(x)]\}\)\(f_{2016}(2)\)\([\underbrace{\left( {1,1, \cdots ,1} \right)}_{n个}]\)\([\overbrace{\left( {1,1, \cdots ,1} \right)}^{n个}]\)

\(\begin{align*} (a+b)^3 &= (a+b)^2(a+b)\\&=(a^2+2ab+b^2)(a+b)\\&=(a^3+2a^2b+ab^2)+(a^2b+2ab^2+b^3)\\&=a^3+3a^2b+3ab^2+b^3\end{align*}\)

临时补充

可以充当段首的文字缩进,

\(\quad\) 一个汉字的间隔 \(\qquad\) 两个汉字的间隔 \(22^{\circ}C\)\({a+1 \over b+1}\)\(\cfrac ab\)\(\bullet\)

一个空格  两个空格  不换行的空格  

\(\eth\) \(\ast\) \(\square\) \(EF\;\;{}_{=}^{//}AH\) 四边形\(\text{▱}ABCD\) \(▱ABCD\)

其他

\[\begin{array}{c|lcr} n & \text{Left} & \text{Center} &\text{Right} \\ \hline 1 & 0.24 & 1 & 125 \\ 2 & -1 & 189 & -8 \\ 3 & -20 & 2000 & 1+10i \\ \end{array} \]

行列式:
\(\left |\begin{array}{cccc} 2&2 \\ 3&3 \end{array}\right |\)

\begin{align}\notag \dot{x}&=\mathbf{A}x+\mathbf{B}u\\ y&=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}x+\begin{bmatrix}1&0\\ 0&1\end{bmatrix}u \end{align}

\[\left (\begin{array}{ccccc} &&a_1&&\\ &a_2&&a_3&\\ a_4&&a_5&&a_6\\ &\cdots&\cdots&\cdots&\\ \end{array}\right )\]

\[\begin{array}{ccccc} a_1&&&&\\ a_2&a_3&a_4&\\ a_5&a_6&a_7&a_8&a_9\\ &\cdots&\cdots&\cdots&\\ \end{array}\]

\(\left (\begin{array}{ccccc} a_1&&&&\\ a_2&a_3&a_4&\\ a_5&a_6&a_7&a_8&a_9\\ &\cdots&\cdots&\cdots&\\ \end{array}\right )\)

$\left (\begin{array}{ccccc} &&a_1&&\\ &a_2&&a_3&\\ a_4&&a_5&&a_6\\ &\cdots&\cdots&\cdots&\\ \end{array}\right )$

\(ab \mathop{\sum\sum\sum}_{a=\frac{1}{2}\times 100000}^{b=\frac{4}{5}}cd\)

\(abx\mathop{\sum\sum\sum}\limits_{a=\frac{1}{2}\times 100000}^{b=\frac{4}{5} }defggas\)

\(\fbox{符号效果只能是单行文本}\)

文本颜色字体 【题组或案例1】已知\(2m+3n=2,m>0,n>0\),求\(\cfrac{4}{m}+\cfrac{1}{n}\)的最小值。

比较理想的习题答案组织方式

脚注:I get 10 times more traffic from \([Google]\)[1] than from [Yahoo][2] or [MSN][3].

文本尺寸

放大及缩小 \(\large\cfrac{2}{3}\) \(\small\cfrac{2}{3}\) \(\huge\kappa\)

\(\tiny\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\scriptsize\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\small\displaystyle\int f^{-1}\;(x-x_a)\;dx\)

\(\normalsize\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\large\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\Large\displaystyle\int f^{-1}\;(x-x_a)\;dx\)

\(\LARGE\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\huge\displaystyle\int f^{-1}\;(x-x_a)\;dx\) \(\hspace{1cm}\) \(\Huge\displaystyle\int f^{-1}\;(x-x_a)\;dx\)

图片

左对齐:
\(\left\{\begin{array}{l}{\frac{4x}{3}+t+4<0}\\{ \frac{2x}{3}+t>0}\\{2x+3>0}\\{3x+2y-4z<0}\end{array}\right.\)

右对齐:

\(\left\{\begin{array}{r}{\frac{4x}{3}+t+4<0}\\{ \frac{2x}{3}+t>0}\\{2x+3>0}\\{3x+2y-4z<0}\end{array}\right.\)

中间对齐:
\(\left\{\begin{array}{c}{\frac{4x}{3}+t+4>0}\\{ \frac{2x}{3}+t<0}\end{array}\right.\)

\(a^{\prime}\) $ \overbrace{AB}$ \(\overleftrightarrow{AB}\) \(\mathring{g}\)
\(\overline{AB}\) \(\widetilde{ac}\)
\(⎰⎱\) \(\lt\gt\) \(\lvert\rvert\) \(\lVert\rVert\) \(\backslash\)

\((\big(\Big(\bigg(\Bigg(\)

\(\begin{matrix} a & b \\ c & d \end{matrix}\) \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

\(\begin{vmatrix} a & b \\ c & d \end{vmatrix}\) \(\begin{Vmatrix} a & b \\ c & d \end{Vmatrix}\) \(\begin{Bmatrix} a & b \\ c & d \end{Bmatrix}\)

\(\def\arraystretch{1.5} \begin{array}{c|c|c} a & b & c \\ \hline d & e & f \\ \hdashline g & h & i \end{array}\)

\(\begin{aligned} a&=b+c \\ d+e&=f \end{aligned}\)

\(\begin{alignedat}{2} 10&x+ &3&y = 2 \\ 3&x+&13&y = 4 \end{alignedat}\)

\(\begin{gathered} a=b \\ e=b+c \end{gathered}\) \(x = \begin{cases} a &\text{if } b \\ c &\text{if } d \end{cases}\)

\(\overbrace{a+b+c}^{\text{note}}\) \(\underbrace{a+b+c}_{\text{note}}\) \(\boxed{\pi=\frac c d}\)

\(\left(x^{\smash{2}}\right)\)

\(\sqrt{\smash[b]{y}}\) \(\sqrt{y}\) \(\vphantom{content}\) \(_u^o\)

\(\stackrel{//}{=}\) \(\overset{!}{=}\) \(\underset{!}{=}\)

\(lim\) \(\lim\)

\(\overbrace{3:2}\)

\(\left\{\begin{aligned} & f(x)=ax^2+bx+c && \text{cases 1}\\ & \sum_{i=1}^na_i && \text{cases 2}\\ & \begin{aligned} g(x)&=cx^2+bx+a\\ &=c_1x^2+b+a_1 \end{aligned} && \text{XXX}\\ & \int_a^bf(x)dx &&\text{f**k} \end{aligned}\right.\)

\(\xrightarrow[under]{over}\) \(\xrightarrow{over}\) \(\xleftarrow[123]{abc}\)

\(\xtwoheadrightarrow{abc}\) \(\xtwoheadleftarrow{abc}\) \(\xmapsto{abc}\)

\(\xlongequal[132]{abc}\) \(\xtofrom[123]{abc}\)

\(\color{blue}{F=ma}\) \(\color{#228B22}{F=ma}\)

\(\checkmark\) \(\surd\)

\(\diagdown\) \(\diagup\)

\(甲\underline{3:2}乙\)

\(\overleftrightarrow{AB}\)

\[\require{AMScd} \begin{CD} f(x)单增\quad\quad @>{勿忘等号}>> f'(x)\geqslant 0\quad\quad @>{\left\{\begin{array}{l}{x=1}\\{y=1}\\{z=3}\end{array}\right.}>> 得到范围^{①111} \\ @V{并列}VV @VV{并列}V \\ f(x)单减\quad\quad @>{勿忘等号}>> f'(x)\leqslant 0\quad\quad @>{\text{验证,排除常函数}}>> 得到范围 \\ \end{CD} \]

\[\require{enclose} \begin{array}{rl} \verb|\enclose{horizontalstrike}{x+y}| & \enclose{horizontalstrike}{x+y} \\ \verb|\enclose{verticalstrike}{\frac xy}| & \enclose{verticalstrike}{\frac xy} \\ \verb|\enclose{updiagonalstrike}{x+y}| & \enclose{updiagonalstrike}{x+y} \\ \verb|\enclose{downdiagonalstrike}{x+y}| & \enclose{downdiagonalstrike}{x+y} \\ \verb|\enclose{horizontalstrike,updiagonalstrike}{x+y}| & \enclose{horizontalstrike,updiagonalstrike}{x+y} \\ \end{array} \]

\[\require{AMScd} \begin{CD} A @>a>> B \\ @V b V V\# @VV c V \\ C @>>d> D \\ \end{CD} \]

\[\frac 12,\frac 1a,\frac a2 \quad \mid \quad \text{2 letters only:} \quad \frac 12a \,, k\frac q{r^2} \]

\[f(x_1,x_2,\underbrace{\ldots}_{\rm ldots} ,x_n) = x_1^2 + x_2^2 + \underbrace{\cdots}_{\rm cdots} + x_n^2 \]

\(\cfrac ab\)

\(y_{_{_N}}\)

\(\sqrt{1+\sqrt[^p\!]{1+\sqrt[^q\!]{1+a}}}\) \(\sqrt{1+\sqrt[^p]{1+\sqrt[^q]{1+a}}}\)

\(f(x)_{\mbox{极大值}}\)

\(\sum\limits_{k=1}^n\)\(\sum\nolimits_{k=1}^n\)

\(\max\limits_{-1\leqslant x\leqslant 2}\)

\[\int_0^1 {x^2} \,{\rm d}x \]

\[\sum_{i=1}^n \frac{1}{i^2} \quad and \quad \prod_{i=1}^n \frac{1}{i^2} \quad and \quad \bigcup_{i=1}^{2} \Bbb{R} \]

\(\rm{SrO+V^{''}_{Sr} \overset{H_2}{\underset{1300℃}{\Longleftrightarrow}} Sr^{\times}_{Sr}+2e^{'}+\frac 12O_2(g)}\)

\(\implies{123}\)

\(\scr WH\)

\[f\left( \left[ \frac{ 1+\left\{x,y\right\} }{ \left( \frac xy + \frac yx \right) (u+1) }+a \right]^{3/2} \right) \tag {行标} \]

\[\require{enclose} \enclose{horizontalstrike}{x+y} \enclose{updiagonalstrike}{x+y} \]

\(\require{enclose}\enclose{updiagonalstrike,downdiagonalstrike}{x+y}\)

\[\require{enclose} \enclose{box}{ \begin{array}{c} f(\top),\, f^2(\top),\, f^3(\top) \,\cdots\, f^n(\top) \\ f(\bot),\, f^2(\bot),\, f^3(\bot) \,\cdots\, f^n(\bot) \\ \end{array} } \]

\[\require{enclose} \begin{array}{c} \enclose{circle}{f(\top),\, f^2(\top),\, f^3(\top) \,\cdots\, f^n(\top)} \\ \enclose{roundedbox}{f(\bot),\, f^2(\bot),\, f^3(\bot) \,\cdots\, f^n(\bot)} \\ \end{array} \]

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