Basta adicionar o script abaixo no cabeçalho do código HTML do seu modelo (Modelo > Editar HTML).
<script type='text/x-mathjax-config'>
// <![CDATA[
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']],
displayMath: [['\\[','\\]'], ['$$','$$']]}});
blogger.ui().viewType_.prototype.onRenderComplete=function(){
MathJax.Hub.Queue(['Typeset',MathJax.Hub])
};
// ]]>
</script>
<!-- script src='http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML' type='text/javascript'>
</script -->
<script src='http://cdn.mathjax.org/mathjax/latest/MathJax.js' type='text/javascript'>
MathJax.Hub.Config({
HTML: ["input/TeX","output/HTML-CSS"],
TeX: { extensions: ["AMSmath.js","AMSsymbols.js"],
equationNumbers: { autoNumber: "AMS" } },
extensions: ["tex2jax.js"],
jax: ["input/TeX","output/HTML-CSS"],
tex2jax: { inlineMath: [ ['$','$'], ["\\(","\\)"] ],
displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
processEscapes: true },
"HTML-CSS": { availableFonts: ["TeX"],
linebreaks: { automatic: true } }
});
</script>
Exemplos:
<script type='text/x-mathjax-config'>
// <![CDATA[
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']],
displayMath: [['\\[','\\]'], ['$$','$$']]}});
blogger.ui().viewType_.prototype.onRenderComplete=function(){
MathJax.Hub.Queue(['Typeset',MathJax.Hub])
};
// ]]>
</script>
<!-- script src='http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML' type='text/javascript'>
</script -->
<script src='http://cdn.mathjax.org/mathjax/latest/MathJax.js' type='text/javascript'>
MathJax.Hub.Config({
HTML: ["input/TeX","output/HTML-CSS"],
TeX: { extensions: ["AMSmath.js","AMSsymbols.js"],
equationNumbers: { autoNumber: "AMS" } },
extensions: ["tex2jax.js"],
jax: ["input/TeX","output/HTML-CSS"],
tex2jax: { inlineMath: [ ['$','$'], ["\\(","\\)"] ],
displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
processEscapes: true },
"HTML-CSS": { availableFonts: ["TeX"],
linebreaks: { automatic: true } }
});
</script>
Exemplos:
$x^3\equiv1\pmod{63}$
$mdc(a,m)=1 \Rightarrow a^{\phi(m)} \equiv 1 \pmod{m}$ (Teo. de Euler)
$\int_0^\infty{e^{-2t}sen(t)\,dt}$
Fonte: MathJax
$mdc(a,m)=1 \Rightarrow a^{\phi(m)} \equiv 1 \pmod{m}$ (Teo. de Euler)
$\int_0^\infty{e^{-2t}sen(t)\,dt}$
Fonte: MathJax
meu teste
ResponderExcluir$\int_{0}^{\pi}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}dx =\frac{22}{7}-\pi$
Ainda não havia testado isto nos comentários. Legal que tenha funcionado! Obrigado.
Excluir