MathJaxでLatexを埋め込む 2018最新版

    >


<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<title>Mathedemo</title>
<script type="text/x-mathjax-config">
  MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}});
</script>
<script type="text/javascript"
  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
</head>

<body>
<h2>Math in TeX notation </h2>

When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
$$ \begin{array}{rcll}
y & = & x^{2}+bx+c\\
  & = & x^{2}+2\times\dfrac{b}{2}x+c\\
  & = & \underbrace{x^{2}+2\times\dfrac{b}{2}x+\left(\frac{b}{2}\right)^{2}}-
      {\left(\dfrac{b}{2}\right)^{2}+c}\\
  &  & \qquad\left(x+{\dfrac{b}{2}}\right)^{2}\\
  & = & \left(x+\dfrac{b}{2}\right)^{2}-\left(\dfrac{b}{2}\right)^{2}+c
  & \left|+\left({\dfrac{b}{2}}\right)^{2}-c\right.\\
    y+\left(\dfrac{b}{2}\right)^{2}-c & = & \left(x+
    \dfrac{b}{2}\right)^{2} & \left|\strut(\textrm{vertex form})\right.\\
y-y_{S} & = & (x-x_{S})^{2}\\
S(x_{S};y_{S}) & \,\textrm{or}\,
    & S\left(-\dfrac{b}{2};\,\left(\dfrac{b}{2}\right)^{2}-c\right)
\end{array} $$

</body>
</html>


お困りですか?この記事で紹介していることをマンツーマンで指導、解説、代行します。まずはお気軽にお問い合わせください。


Close Menu