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\begin{center}\section*{\textcolor{red}{Correction}}\end{center}

\subsection{}
Soit l'équation $\dfrac{5x-4}{3}-\dfrac{7x+5}{12}=\dfrac{3x-7}{2}+\dfrac{x-2}{6}$.\\
On multiplie tout par 12 : on obtient :\\

$4(5x-4)-7x-5=6(3x-7)+2(x-2)\\
\Leftrightarrow 20x-16-7x-5=18x-42+2x-4\\
\Leftrightarrow 20x-7x-1x-2x=16+5-42-4\\
\Leftrightarrow -7x=-25\Leftrightarrow x=\dfrac{25}{7}$.\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{\dfrac{25}{7}\right\}}}$


\subsection{}


\begin{enumerate}
\item $7(2-5x)(3x+1)=0=0 \Leftrightarrow 2-5x=0 ou 3x+1=0$
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-\dfrac{1}{3}~;~\dfrac{2}{5}\right\}}}$


\item $(x+4)(5x-3)-2(x+3)(x+4)=0 \text{On factorise par }(x+4) :\\
(x + 4)[(5x - 3) - 2(x + 3)] = 0\Leftrightarrow (x + 4)(5x - 3 - 2x - 6) = 0\Leftrightarrow (x + 4)(3x - 9) = 0\Leftrightarrow 3(x + 4)(x - 3) = 0\Leftrightarrow  x+4=0\text{ ou }x-3=0\\
\Leftrightarrow x = -4\text{ ou }x = 3$\\
$\boxed{\textcolor{red}{\mathscr{S} = \{-4~;~3\}}}$

\item $(3x+1)^2 =(4-x)(3x+1)\text{ On annule le second membre et on factorise par }3x+1 \\
(3x+1)^2 -(4-x)(3x+1)=0\Leftrightarrow (3x + 1)[(3x + 1) - (4 - x)] = 0\Leftrightarrow (3x + 1)(3x + 1 - 4 + x) = 0\Leftrightarrow (3x + 1)(4x - 3) = 0\\
 \Leftrightarrow 3x+1=0\text{ ou }4x-3=0$\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-\dfrac{1}{3}~;~\dfrac{3}{4}\right\}}}$

\item $x(x-2)+(x-2)-(x-2)(7x+9)=0\text{ On factorise par }x-2\\
\Leftrightarrow (x - 2)[x + 1 - (7x + 9)] = 0\Leftrightarrow (x - 2)(x + 1 - 7x - 9) = 0\Leftrightarrow (x - 2)(-6x - 8) = 0\Leftrightarrow  -2(x - 2)(3x + 4) = 0\\
\Leftrightarrow x-2=0\text{ ou }3x+4=0\Leftrightarrow x = 2\text{ ou }x = -\dfrac{4}{3}$\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-\dfrac{4}{3}~;~2\right\}}}$

\item $(7x + 3)^2 = (x - 1)^2\Leftrightarrow \left(7x + 3\right)^2 -\left( x - 1\right)^2=0\Leftrightarrow \left[(7x+3)+(x-1)\right]\left[(7x+3)-(x-1)\right]=0\Leftrightarrow (8x+2)(6x+4)=0\\\
\Leftrightarrow 2(4x+1)\times 2(3x+2)=0\Leftrightarrow 4(4x+1)(3x+2)=0\Leftrightarrow 4x+1=0\text{ ou }3x+2=0$\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-\dfrac{2}{3}~;~-\dfrac{1}{4}\right\}}}$

\item $4x^2 -25-(2x+5)(7-x)=0 \Leftrightarrow (2x - 5)(2x + 5) - (2x + 5)(7 - x) = 0\Leftrightarrow (2x + 5)[(2x - 5) - (7 - x)] = 0\\
\Leftrightarrow  (2x + 5)(3x - 12) = 0\Leftrightarrow 3(2x + 5)(x - 4) = 0\Leftrightarrow   2x+5=0\text{ ou }x-4=0
$\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-\dfrac{5}{2}~;~4\right\}}}$
\end{enumerate}


\subsection{}

\begin{enumerate}
\item 
\begin{enumerate}
\item $E(x)=9x^2 -24x+16-\left(4x^2 -20x+25\right) =9x^2 -24x+16-4x^2 +20x-25 = \boxed{\textcolor{red}{5x^2 - 4x - 9}}$

\item $E(x)=(3x-4)^2 -(2x-5)^2=[(3x-4)+(2x-5)][(3x-4)-(2x-5)]=\boxed{\textcolor{red}{(5x-9)(x+1)}}$
\end{enumerate}


\item 
\begin{enumerate}
\item $E(x)=0\Leftrightarrow (5x-9)(x+1)\Leftrightarrow 5x-9=0\text{ ou }x+1=0.$\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-1~;~\dfrac{9}{5}\right\}}}$

\item $E(x)=-9\Leftrightarrow 5x^2 -4x-9=-9\Leftrightarrow x(5x-4)=0\Leftrightarrow x=0\text{ ou }5x-4=0$.\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{0~;~\dfrac{4}{5}\right\}}}$

\item $E(x)=(2x-5)(x+1)\Leftrightarrow (x+1)(5x-9)-(2x-5)(x+1)=0 \Leftrightarrow (x + 1)(5x - 9 - 2x + 5) = 0\\
\Leftrightarrow (x + 1)(3x - 4) = 0$.\\
$\boxed{\textcolor{red}{\mathscr{S}=\left\{-1~;~\dfrac{4}{3}\right\}}}$

\end{enumerate}

\end{enumerate}

\label{fin}
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