根号下的 \strut 的影响可以减弱吗？

Question

\strut 对平方根下的垂直间距的影响非常严重。我希望效果约为原来的 20%。否则，strut 可以解决很多问题。

% without \strut
\noindent \hspace{18pt}\large$\bigg[\hspace{1.4pt}x\hspace{1.4pt}'\hspace{-0.6pt}+\dfrac{\big(\hspace{0.4pt}9\hspace{-0.6pt}-\hspace{-0.6pt}5\hspace{0.4pt}\surd{\hspace{1.0pt}2}\hspace{0.8pt}\big) \hspace{1.2pt}\sqrt{  \hspace{1.2pt}4\hspace{-0.6pt}+\hspace{-0.6pt}2\hspace{0.4pt}\surd{\hspace{1.0pt}2}}}{62}\hspace{0.8pt}\bigg]^{\hspace{0.8pt}2}$

% with \strut
\vspace{6pt}
\noindent \hspace{18pt}$\bigg[\hspace{1.4pt}x\hspace{1.4pt}'\hspace{-0.6pt}+\dfrac{\big(\hspace{0.4pt}9\hspace{-0.6pt}-\hspace{-0.6pt}5\hspace{0.4pt}\surd{\hspace{1.0pt}2}\hspace{0.8pt}\big) \hspace{1.2pt}\sqrt{\strut\hspace{1.2pt}4\hspace{-0.6pt}+\hspace{-0.6pt}2\hspace{0.4pt}\surd{\hspace{1.0pt}2}}}{62}\hspace{0.8pt}\bigg]^{\hspace{0.8pt}2}$

Answer 1

您必须尝试使用​​小一点的 \strut。如您在以下示例中所见，0.1pt 的差异可能会导致尺寸发生相当大的变化：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
 $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\ht\strutbox}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.1pt}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.2pt}(x_2-x_1)^2+(y_2-y_1)^2}$.

 $\sqrt{\rule{0pt}{\dimexpr \ht\strutbox+0.3pt}(x_2-x_1)^2+(y_2-y_1)^2}$.
\end{document}