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\begin{document}


\begin{center}\section*{\textcolor{red}{correction du TD \no 11 (dÃ©veloppements )}}\end{center}




\subsection{}
%Simplifier $a(b-c)+b(c-a)+c(a-b)$ pour des nombres $a$, $b$ et $c$ quelconques.
$a(b-c)+b(c-a)+c(a-b)=ab-ac+bc-ba+ca-cb=ab-ac+bc-ab+ac-bc=\boxed{\textcolor{red}{0}}$

\subsection{}
DÃ©velopper les expressions suivantes :


\begin{align*}
A(x)&amp;=\Distri{0}{3}{7}{5}\\
A(x)&amp;=\Distri[Etape=2]{0}{3}{7}{5}\\
A(x)&amp;=\boxed{\textcolor{red}{\Distri[Etape=3]{0}{3}{7}{5}}}
\end{align*}

%\bigskip

\begin{align*}
B(x)&amp; =\Distri{0}{9}{3}{-7}\\
B(x)&amp;=\Distri[Etape=2]{0}{9}{3}{-7}\\
B(x)&amp;=\Distri[Etape=3]{0}{9}{3}{-7}\\
B(x)&amp;=\boxed{\textcolor{red}{27x-63}}
\end{align*}

%\bigskip

\begin{align*}
C(x)&amp;=\Distri{0}{-2}{3}{-5}\\
C(x)&amp;=\Distri[Etape=2]{0}{-2}{3}{-5}\\
C(x)&amp;=\Distri[Etape=3]{0}{-2}{3}{-5}\\
C(x)&amp;=\boxed{\textcolor{red}{-6x+10}}
\end{align*}

%\bigskip

\begin{align*}
D(x)&amp;=\Distri{1}{1}{2}{3}\\
D(x)&amp;=\Distri[Etape=2]{1}{1}{2}{3}\\
D(x)&amp;=\boxed{\textcolor{red}{\Distri[Etape=3]{1}{1}{2}{3}}}
\end{align*}

%\bigskip

\begin{align*}
E(x)&amp;=\Distri{2}{8}{1}{5}\\
E(x)&amp;=\Distri[Etape=2]{2}{8}{1}{5}\\
E(x)&amp;=\Distri[Etape=3]{2}{8}{1}{5}\\
E(x)&amp;=\boxed{\textcolor{red}{\Distri[Etape=4]{2}{8}{1}{5}}}
\end{align*}

%\bigskip

\begin{align*}
F(x)&amp;=\Distri{5}{4}{4}{7}\\
F(x)&amp;=\Distri[Etape=2]{5}{4}{4}{7}\\
F(x)&amp;=\Distri[Etape=3]{5}{4}{4}{7}\\
F(x)&amp;=\boxed{\textcolor{red}{\Distri[Etape=4]{5}{4}{4}{7}}}
\end{align*}

%\bigskip

\begin{align*}
G(x)&amp;=\Distri{7}{-4}{2}{1}\\
G(x)&amp;=\Distri[Etape=2]{7}{-4}{2}{1}\\
G(x)&amp;=\Distri[Etape=3]{7}{-4}{2}{1}\\
G(x)&amp;=\boxed{\textcolor{red}{\Distri[Etape=4]{7}{-4}{2}{1}}}
\end{align*}

%\bigskip

\begin{align*}
H(x)&amp;=\Distri[RAZ]{3}{1}{4}{2}-\Distri{0}{5}{3}{-2}\\
H(x)&amp;=\Distri[Etape=2]{3}{1}{4}{2}-(\Distri[Etape=2]{0}{5}{3}{-2})\\
H(x)&amp;=\Distri[Etape=3]{3}{1}{4}{2}-(\Distri[Etape=3]{0}{5}{3}{-2})\\
H(x)&amp;=\Distri[Somme,Etape=4]{3}{1}{4}{2}-(\Distri[Difference,Etape=4]{0}{5}{3}{-2})\\
H(x)&amp;=\boxed{\textcolor{red}{\Resultat}}
\end{align*}

%\bigskip

\begin{align*}
I(x)&amp;= \Distri[RAZ]{5}{-4}{3}{7}-\Distri{4}{-3}{5}{9}\\
I(x)&amp;= \left[\Distri[Etape=2]{5}{-4}{3}{7}\right]-\left[\Distri[Etape=2]{4}{-3}{5}{9}\right]\\
I(x)&amp;= \left[\Distri[Etape=3]{5}{-4}{3}{7}\right]-\left[\Distri[Etape=3]{4}{-3}{5}{9}\right]\\
I(x)&amp;= \left[\Distri[Somme,Etape=4]{5}{-4}{3}{7}\right]-\left[\Distri[Difference,Etape=4]{4}{-3}{5}{9}\right]\\
I(x)&amp;=\boxed{\textcolor{red}{\Resultat}}
\end{align*}

\begin{multicols}{2}
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\subsection{}
ComplÃ©ter \dots par des nombres entiers :

\begin{enumerate}[1)]

%\bigskip

\item $ (5x+6)(2x+3) = 10 x^2 +27 x+18$

%\bigskip

\item $ (4x+3)(x+5) =4 x^2 +23 x+15 $

%\bigskip

\item $(2x -6)(3x+4) = 6 x^2- 10 x-24$

%\bigskip

\item $3x(4 x+2)+5 x = 12x^2 +11x$
\end{enumerate}

\subsection{}
%Rappeler le dÃ©veloppement de :
\begin{enumerate}[$\bullet$]
\item $(a+b)^2=a^2+2ab+b^2$

\item $(a-b)^2=a^2-2ab+b^2$

\item $(a+b)(a-b)=a^2-b^2$
\end{enumerate}

\subsection{}
%DÃ©velopper, en utilisant les identitÃ©s remarquables :




\begin{align*}
A(x)&amp;= \Distri[Remarquable]{2}{3}{}{}\\
A(x)&amp;= \Distri[Remarquable,Etape=2]{2}{3}{}{}\\
A(x)&amp;=\boxed{\textcolor{red}{ \Distri[Remarquable,Etape=3]{2}{3}{}{}}}
\end{align*}

\begin{align*}
B(x)&amp;= \Distri[Remarquable]{3}{-5}{}{}\\
B(x)&amp;= \Distri[Remarquable,Etape=2]{3}{-5}{}{}\\
B(x)&amp;=\boxed{\textcolor{red}{ \Distri[Remarquable,Etape=3]{3}{-5}{}{}}}
\end{align*}

\begin{align*}
C(x)&amp;= \Distri[Remarquable]{7}{3}{7}{-3}\\
C(x)&amp;= \Distri[Remarquable,Etape=2]{7}{3}{7}{-3}\\
C(x)&amp;=\boxed{\textcolor{red}{ \Distri[Remarquable,Etape=3]{7}{3}{7}{-3}}}
\end{align*}

\begin{align*}
D(x)&amp;= \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\
D(x)&amp;=\sqrt{3}^2-\sqrt{2}^2\\
D(x)&amp;=3-2=\boxed{\textcolor{red}{1}}
\end{align*}


\end{multicols}
\label{fin}
\end{document}  </pre></body></html>