我做了一个公式,它太长了,所以它显示不居中。我希望将它移动到中心并仅为此公式扩展文本宽度,因此在此公式之后,我希望文本宽度恢复为默认值。
\documentclass{article}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}
\usepackage{mathtools}
\begin{document}
\centering
\begin{align}
\label{eq:MgSO4}
\left( \ce{MgSO4}\right)&=-\frac{10^{\text{pH}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{29441213}{5000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}\\
\label{eq:CaSO4}
\left(\ce{CaSO4}\right)&=-\frac{10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}\\
\label{eq:CaMg(CO3)2}
\left(\ce{CaMg(CO3)2}\right)&=-\frac{10^{2 \text{pH}+\frac{58880041}{10000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}
\end{align}
\end{document}
答案1
当然,从风格上来说,这有点不合时宜,但这是可行的:
\documentclass{article}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}
\usepackage{mathtools}
%\UseTblrLibrary{booktabs, siunitx} % throws an error
\usepackage{chngpage}
\begin{document}
%\centering % why is this here?
\begin{adjustwidth}{-2.5cm}{-2.5cm}
\begin{align}
\label{eq:MgSO4}
\left( \ce{MgSO4}\right)&=-\frac{10^{\text{pH}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{29441213}{5000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}\\
\label{eq:CaSO4}
\left(\ce{CaSO4}\right)&=-\frac{10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}\\
\label{eq:CaMg(CO3)2}
\left(\ce{CaMg(CO3)2}\right)&=-\frac{10^{2 \text{pH}+\frac{58880041}{10000000}}}
{
\splitfrac{-10^{\frac{16393 t^2}{25000 (20 t+5463)}+\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t^2+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}
\end{align}
\end{adjustwidth}
\end{document}
chngpage
这将使用与环境相关的包adjustwidth
。
答案2
一种可能的解决方案是,结合加载geometry
,以获得更合适的边距,fleqn
环境和来自的中等大小的分数nccmath
,以及\mathrlap
来自的mathtools
:
\documentclass{article}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}
\usepackage{nccmath, mathtools}
\usepackage[showframe]{geometry}
\begin{document}
\begin{fleqn}
\begin{align}
\label{eq:MgSO4}
\left( \ce{MgSO4}\right)&=-\mathrlap{\mfrac{10^{\text{pH}+\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{29441213}{5000000}}}
{
\splitfrac{-10^{\frac{16393 t²}{25000 (20 t+5463)}+\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}}\\
\label{eq:CaSO4}
\left(\ce{CaSO4}\right)&=-\mathrlap{\mfrac{10^{\frac{16393 t²}{25000 (20 t+5463)}+\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}
{
\splitfrac{-10^{\frac{16393 t²}{25000 (20 t+5463)}+\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}}\\
\label{eq:CaMg(CO3)2}
\left(\ce{CaMg(CO3)2}\right)&=-\mathrlap{\mfrac{10^{2 \text{pH}+\frac{58880041}{10000000}}}
{
\splitfrac{-10^{\frac{16393 t²}{25000 (20 t+5463)}+\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{89554959 t}{500000 (20 t+5463)}+\frac{340471}{5 (20 t+5463)}+\frac{477}{2000000}}}{-10^{\text{pH}+\frac{58880041}{10000000}}
\left(10^{\text{pH}}+10^{\frac{47580 t²+12996477 t+5804780000}{100000 (20 t+5463)}+\frac{477}{2000000}}\right)}}}
\end{align}
\end{fleqn}
\end{document}
答案3
增加文本块的宽度可能会略有帮助。但是,我认为最好将重点放在识别三个方程的分子和分母项中的共同“块”或“块”。例如,人们不仅可能注意到所有三个分母都相同,而且第二个方程中分数的分子出现在所有三个大分母中。因此,第二个方程的 RHS 可以写为\frac{U}{U+V}
。在收集更多共同术语后,人们得出以下结论:
\documentclass{article}
\usepackage[version=4]{mhchem}
\usepackage{mathtools}
\DeclareMathOperator{\ten}{ten}
\newcommand\pH{\mathrm{pH}}
\begin{document}
\noindent
Put $\ten(x)\equiv 10^{x}$. Then
\begin{align}
( \ce{MgSO4}) &= A/D \label{eq:MgSO4} \\
(\ce{CaSO4}) &= B/D \label{eq:CaSO4} \\
(\ce{CaMg(CO3)2})&= C/D \label{eq:CaMg(CO3)2}
\end{align}
where
\begin{align*}
A&=\ten(\pH+H+N) \\
B&=\ten(G+H+J+K+L) \\
C&=\ten(2\pH+M) \\
D&=B + \ten(\pH+M) [\ten(\pH)+\ten(H+L)] \\
\shortintertext{and}
G&= \frac{16393 t^2}{25000 (20t+5463)} \\[\jot]
H&= \frac{47580 t^2+12996477 t+5804780000}{100000 (20t+5463)} \\[\jot]
J&= \frac{89554959 t}{500000 (20t+5463)} \\[\jot]
K&= \frac{340471}{5 (20t+5463)} \\[\jot]
L&= \frac{477}{2000000} \\[\jot]
M&= \frac{58880041}{10000000} \\[\jot]
N&= \frac{29441213}{5000000} \,.
\end{align*}
\end{document}