\documentclass{article}
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\begin{document}

\begin{enumerate}
	\item 
Soit $\frac{3X+7}{(X+1)^2}\in \R(X)$.

Posons:
\begin{align*}
	\frac{3X+7}{(X+1)^2} &= \frac{A}{X+1} + \frac{B}{(X+1)^2} \\
\end{align*}

On a:
\begin{align*}
	\frac{3X^2+7X}{(X+1)^2} &= \frac{AX}{X+1} + \frac{BX}{(X+1)^2} \\
	X \to \infty \implies \frac{3}{1} &= \frac{A}{1} + 0 \\
					  &\implies \fbox{A=3}
\end{align*}

Et:

\begin{align*}
	X = 0 \implies \frac{7}{1} &= \frac{3}{1} + \frac{B}{1} \\
				   &\implies \fbox{B=4}
\end{align*}

D'où \fbox{$f = x\mapsto \frac{3}{x+1} + \frac{4}{(x+1)^2}$}.

\item On a \[
		f = 3  \cdot  \left( x\mapsto \frac{1}{1+x} \right) + 4  \cdot \left( x\mapsto \frac{1}{1+x} \right)^2
\] 

Or $x\to 0$, donc on effectue le $\text{DL}_n(0)$ de $\frac{1}{1+\operatorname{id}}$, puis, par linéarité:

\begin{align*}
	f(x) &= 3 \sum_{k=0}^{n} (-1)^{k}x^{k} + 4 \left( \sum_{k=0}^{n} (-1)^{k}x^{k} \right)^2 + o(x^{n}) \\
		&=   \\
\end{align*}

\item Particularisons en $n=3$.
\end{enumerate}



\end{document}
