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              pdftitle={Exercices d'entrainement sur les équations},%
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\begin{document}
 \titre{Équations}
 \classe{\prems}
 \typedoc*{Entrainement}{4}
 \date{}
 
\affichetitre

Toutes ces équations et inéquations peuvent se ramener à des problèmes du premier degré.

\Opensolutionfile{ans}[ans1]

\begin{exercice}
Pour chacune des équations et des inéquations suivantes :
\begin{enumerate}
\item Donner le plan de résolution
\item Résoudre l'(in)équation
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}[label=$(P_{\arabic*}) :$,widest=10,itemsep=10pt]
\item $(x - 5)(2x + 3) + 4(x - 5) = 0$
\begin{sol}
$\cals=\left\{-\dfrac{7}{2};5\right\}$
\end{sol}

\item $(x - 5)(2x + 3) - 2x(x + 1) = 0$
\begin{sol}
$\cals=\left\{-\dfrac{5}{3}\right\}$
\end{sol}

\item $x^2 + 3x = 5x$
\begin{sol}
$\cals=\left\{0;2\right\}$
\end{sol}

\item $x^2 + 3x = 3x + 9$
\begin{sol}
$\cals=\left\{-3;3\right\}$
\end{sol}


%\item $(x - 1)(2x + 3) + (x - 1)(3x - 4) = 1$

\item $(4x + 1)(x + 2) - 2x(2x + 3) = 0$
\begin{sol}
$\cals=\left\{-\dfrac{2}{3}\right\}$
\end{sol}
\item $(x-4)(2x+3)+(x-4)(x-7) \geqslant 0$ ;
\begin{sol}
$\cals=\intervalleOF{-\infty}{\dfrac{4}{3}}\cup \intervalleFO{4}{+\infty}$
\end{sol}
\item $(2x + 1)^2 - 3x \leqslant 4x(x + 2) - 3$
\begin{sol}
$\cals=\intervalleFO{\dfrac{4}{7}}{+\infty}$
\end{sol}

\item $(x+1)^2 \leqslant (2x-3)^2$ ;
\begin{sol}
$\cals=\intervalleOF{-\infty}{\dfrac{2}{3}}\cup \intervalleFO{4}{+\infty}$
\end{sol}
\item $\dfrac{3x^2-4x-3}{x^2-4}\leqslant3$ ;
\begin{sol}
$\cals=\intervalleOO{-2}{2} \cup \intervalleFO{\dfrac{9}{4}}{+\infty}$
\end{sol}
\item $(x^2 - 9)(x + 1) + (x + 3)(x^2 - 1) = 0$
\begin{sol}
$\cals=\left\{-3;-1;2\right\}$
\end{sol}

   \item $\dfrac{x^2+4x-3}{x^2-1}=1$
\begin{sol}
$\cals=\left\{\dfrac{1}{2}\right\}$
\end{sol}
   
   \item $\dfrac{x^2}{x-1}=1+\dfrac{1}{x-1}$
\begin{sol}
$\cals=\left\{0\right\}$
\end{sol}
   
   
   \item $\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{3}{x(x+3)}$
\begin{sol}
$\cals=\varnothing$
\end{sol}

%\item $(2x-5)(-x-3) < 15$ ;
\item {\scriptsize $(-x+1)(6x-5)(x+3)+(-x+1)(6x-5)(x-5) > 0$ ;}
\begin{sol}
$\cals=\intervalleOO{-\infty}{\dfrac{5}{6}}$
\end{sol}
\item $(2x+1)^2(x^2+1)-9(x^2+1)<0$.
\begin{sol}
$\cals=\intervalleOO{-2}{1}$
\end{sol}
%\item $\dfrac{3x-1}{2-x}>0$ ;
%\item $\dfrac{4x-7}{3x+2}\leqslant4$ ;
\item $\dfrac{-2x+x^2}{5-3x}>0$ ;
\begin{sol}
$\cals=\intervalleOO{-\infty}{0} \cup \intervalleFO{\dfrac{5}{3}}{2}$
\end{sol}
\item $x^2(2x + 1) = 4(2x + 1)$
\begin{sol}
$\cals=\left\{-2;-\dfrac{1}{2};2\right\}$
\end{sol}
\item $x^2 + 4x + 4 = (x + 2)(2x - 1)$
\begin{sol}
$\cals=\left\{-2;3\right\}$
\end{sol}


\item $(x - 1)(2x + 3) + (x - 1)(3x - 4) = 0$
\begin{sol}
$\cals=\left\{1;\dfrac{1}{5}\right\}$
\end{sol}
\item $\dfrac{x^2-2x+1}{-3x+2}>0$.
\begin{sol}
$\cals=\intervalleOO{-\infty}{\dfrac{2}{3}}$
\end{sol}
%   \item $\dfrac{2x+3}{x-2}=\dfrac{x+3}{x-1}$


\end{enumerate}
\end{multicols}
\Closesolutionfile{ans}

\textbf{Réponses}

\begin{multicols}{2}
\input{ans1}
\end{multicols}
\end{exercice} 
\end{document}