化学方程式中的长等号怎么表示呢？
怎么让等号的上下输入字符串呢？
$\Longleftarrow$

$CO_3^-$
$\_$

\begin{align} & K=5A \\ & C=25 \end{align}
\begin{align}
\{ \sqrt{\frac{a^2+b^2}2}&\ge \frac{a+b}{2}\\
&\ge \sqrt{ab}\\
&\ge \frac2{\frac1a+\frac1b}
\end{align}
$\left.{\left \{ (z+0.4)\frac { H(z) }{ z }\right\} } \right|_{ z=-0.4 }$
$\left \{ \begin{aligned} & a = b \\ & c = d \end{aligned} \right.$
$\left \{
\begin{aligned}
a = b \\
c = d
\end{aligned}
\right\}$
$\left [
\begin{aligned}
a = b \\
c = d
\end{aligned}
\right ]$

$|x-1|={[sqrt{2x}k]}/2$在区间[k-1,k+1] 上有不等两实数根，k的取值范围

$|x-1|={[sqrt{2x}k]}/2$在区间[k-1,k+1] 上有不等两实数根，k的取值范围
xwh19771018 发表于 2012-4-30 11:30

仔细看另一个置顶贴的代码输入。
打好了再去数学讨论区提问题吧。

$f(x)=4x^3+3tx^2-6t^2x+t-1$

洛仑兹坐标变换公式

\[
\left\{
\begin{array}{l}
x = \frac{x'+vt'}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \\
y = y' \\
z = z' \\
t = \frac{t'+\frac{v}{c^2}x'}{\sqrt{1-\left(\frac{v}{c}\right)^2}}
\end{array}
\right.
\]

Cardano公式：

一元三次方程\(ax^3+bx^2+cx+d=0 \, (a \ne 0)\)的解是：

\begin{align*}
x_1 =
&-\frac{b}{3 a}\\
&-\frac{1}{3 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&-\frac{1}{3 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_2 =
&-\frac{b}{3 a}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_3 =
&-\frac{b}{3 a}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\frac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}
\end{align*}

48# 叶剑飞Victor

根号里面的 1/2 为什么用 \tfrac ？有点小

我也试试
$f(x)=ax^2(x-2)$

$q_3q_4q_3...$, 同时$q_3\models\phi$

$q_3q_4q_3...$, 同时$q_3\models\phi$
孤星赶月 发表于 2012-12-22 13:17

$\models$ 是什么符号？

来个恶心的：


一元四次方程$ax^4+bx^3+cx^2+dx+e=0 (a \ne 0)$的求根公式：

\begin{array}{l}
{x_1=-\frac{b}{4a}+\frac{1}{2} \sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}}-\frac{1}{2} \sqrt{\frac{b^2}{2a^2}-\frac{4c}{3a}-\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}-\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}+\frac{-b^3+4abc-8a^2d}{4a^3\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}} {3\sqrt[3]{2}a}}}}}\\\\

{x_2=-\frac{b}{4a}+\frac{1}{2} \sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}}+\frac{1}{2} \sqrt{\frac{b^2}{2a^2}-\frac{4c}{3a}-\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}-\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}+\frac{-b^3+4abc-8a^2d}{4a^3\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}} {3\sqrt[3]{2}a}}}}}\\\\

{x_3=-\frac{b}{4a}-\frac{1}{2} \sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}}-\frac{1}{2} \sqrt{\frac{b^2}{2a^2}-\frac{4c}{3a}-\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}-\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}-\frac{-b^3+4abc-8a^2d}{4a^3\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}} {3\sqrt[3]{2}a}}}}}\\\\

{x_4=-\frac{b}{4a}-\frac{1}{2} \sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}}+\frac{1}{2} \sqrt{\frac{b^2}{2a^2}-\frac{4c}{3a}-\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}-\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}{3\sqrt[3]{2}a}-\frac{-b^3+4abc-8a^2d}{4a^3\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+\frac{\sqrt[3]{2}\left(c^2-3bd+12ae\right)}{3a\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}}+\frac{\sqrt[3]{2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{-4\left(c^2-3bd+12ae\right)^3+\left(2c^3-9bcd+27ad^2+27b^2e-72ace\right)^2}}} {3\sqrt[3]{2}a}}}}}\\\\

{\Delta=256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2cd^2e-27a^2d^4+144ab^2ce^2-6ab^2d^2e-80abc^2de+18abcd^3+16ac^4e-4ac^3d^2-27b^4e^2+18b^3 c d e-4 b^3 d^3-4 b^2 c^3 e+b^2 c^2 d^2}
\end{array}







好吧，显示得一团糟，还是 http://mathurl.com/c7m26yt 显示得好一些

\[f(x)=\left\{\begin{aligned} &ax+1,&&-1\leqslant x<0,\\&\dfrac{bx+2}{x+1},&&0\leqslant x\leqslant1,\end{aligned}\right.\]

\[f(x)=\left\{\begin{align}&ax+1,&-1\leqslant x<1,\\&\dfrac{bx+2}{x+1},&0\leqslant x\leqslant1,\end{align}\right.\]
\[f(x)=\begin{align}&ax+1,&-1\leqslant x<1,\\&\dfrac{bx+2}{x+1},&0\leqslant x\leqslant1,\end{align}\]

\[f(x)=\begin{cases}&ax+1,&-1\leqslant x<1\\&\frac{bx+2}{x+1},&0\leqslant x\leqslant1\end{cases}\]

54# 李斌斌755
有点意思

还是这样漂亮
\[f(x)=\left\{\begin{aligned} &ax+1,&-1\leqslant x<0,\\&\dfrac{bx+2}{x+1},&0\leqslant x\leqslant1,\end{aligned}\right.\]

\[x^2+y^2=1\]

打开这个页面卡了一阵子，估计是因为53#的公式太长，mathjax 需要点时间才能排好……

赶紧翻个页先

赶紧翻个页先