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\usepackage{amsmath}
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\usepackage[thref,thmmarks,standard,amsmath,hyperref]{ntheorem}
\usepackage[]{hyperref}

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\let\:\overline
\def\cW{\mathrm{W}}

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\theoremstyle{marginbreak}
\newtheorem{thm}{Äá»‹nh lÃ½}

\theoremstyle{plain}
\newtheorem{thm2}[thm]{Má»‡nh Ä‘á»}

\theoremstyle{changebreak}
\newtheorem{thm3}[thm]{Äá»‹nh lÃ½}

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\begin{document}

\section{Má»¥c Ä‘áº§u tiÃªn}
\label{sec:test}

\begin{thm}[Ãnh xáº¡ Weingarten]
Náº¿u $k_1$, $k_2$ lÃ&nbsp; cÃ¡c giÃ¡ trá»‹ riÃªng cá»§a Ã¡nh xáº¡ Weingarten cá»§a $X$,
thÃ¬ $kk_1$, $kk_2$ lÃ&nbsp; cÃ¡c giÃ¡ trá»‹ riÃªng cá»§a Ã¡nh xáº¡ Weingarten cá»§a $\:X$.
ÄÃ¢y lÃ&nbsp; \thmname.
\end{thm}

\begin{thm2}[Ãnh xáº¡ Weingarten]
\label{test}
Náº¿u $k_1$, $k_2$ lÃ&nbsp; cÃ¡c giÃ¡ trá»‹ riÃªng cá»§a Ã¡nh xáº¡ Weingarten cá»§a $X$,
thÃ¬ $kk_1$, $kk_2$ lÃ&nbsp; cÃ¡c giÃ¡ trá»‹ riÃªng cá»§a Ã¡nh xáº¡ Weingarten cá»§a $\:X$.
\end{thm2}

\begin{Proof}
Tháº­t váº­y, náº¿u $v$ lÃ&nbsp; má»™t vÃ©c tÆ¡ riÃªng á»©ng vá»›i giÃ¡ trá»‹ riÃªng $k_1$ cá»§a $\cW$,
thÃ¬ $\cW(v)=k_1v$. Sá»­ dá»¥ng (\ref{eq:W.W}), ta cÃ³
$\:\cW(T^{-1}v) = k\cdot T^{-1}(k_1v)=kk_1\cdot T^{-1}v$.
VÃ¬ $w=T^{-1}v\not=0$, ta cÃ³ $w$ lÃ&nbsp; vÃ©c tÆ¡ riÃªng cá»§a $\cW$ á»©ng vá»›i giÃ¡ trá»‹ riÃªng $kk_1$.
\end{Proof}

\section{Má»¥c tiáº¿p theo}

\begin{thm}[Nhá»‹ thá»©c Newtown]
Newtown Ä‘Ã£ tÃ¬m ra Ä‘Æ°á»£c
\begin{equation}\label{eq:W.W}
(a+b)^n=\sum_{k=0}^n{n\choose k}a^{n-k}b^k
\end{equation}
\end{thm}

\section{Kiá»ƒm tra}

Tham kháº£o má»Ÿ rá»™ng: Xem \thref{test} (cho bá»Ÿi \verb#\thref{test}#)

\bigskip
Tham kháº£o thÆ°á»ng: Xem \ref{sec:test} (má»¥c), \ref{eq:W.W} (phÆ°Æ¡ng trÃ¬nh)

\bigskip
Danh sÃ¡ch Ä‘á»‹nh lÃ½:

\listtheorems{thm,thm2}
\end{document}
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