Addendum:
I've written exercises with a simple box but I want to turn in the picture below style.Here's the code:
\documentclass[11pt]{book}
\usepackage[top=2cm, bottom=2.5cm, left=2.5cm, right=2.5cm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[most]{tcolorbox}
\usepackage{eso-pic}
\usepackage{enumerate}
\usepackage{fancyhdr}
\usepackage{mathrsfs}
\usepackage{cancel}
\usepackage{xcolor}
\usepackage{graphicx}
\usepackage{pgf,tikz,pgfplots}
\usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{tikz,tkz-tab}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usetikzlibrary{arrows}
\usepackage{mdframed}
\usepackage{hyperref}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{tikz,lmodern}
\begin{document}
\section{EXERCICES}
\begin{tcolorbox}[enhanced,breakable,colback=white,colframe=green!50!white,
colbacktitle=white!15!pink,
coltitle=pink!50!black,
borderline={0.5mm}{0mm}{green!15!white},
borderline={0.5mm}{0mm}{green!50!white,dashed},
attach boxed title to top center={yshift=-2mm},
boxed title style={boxrule=0.4pt},
title=EXERCICES]
\underline{\textbf{Exercice 1}}
\\
Simplifier les expressions suivantes:
\begin{enumerate}
\item $A=\dfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
\item $B=\dfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
\item $C=\dfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
\item $D=\dfrac{\left[\left(\dfrac23\right)^2\right]^6\times \left[\left(\dfrac35\right)^{-2}\right]^3\times\left[\left(\dfrac52\right)^2\right]^{-6}}{\left(\dfrac46\right)^6}$
\end{enumerate}
\underline{\textbf{Exercice 2}}
\begin{enumerate}
\item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.\\
Montrer que $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:
\\ $\dfrac{a}{ab+a+1}+\dfrac{b}{bc+c+1}+\dfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:
\\
$\dfrac{4a^2-1}{(a-b)(a-c)}+\dfrac{4b^2-1}{(b-c)(b-a)}+\dfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}
\underline{\textbf{Exercice 3}}
\\
\\
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\dfrac{a}{b}=\dfrac{c}{d}$.
\begin{enumerate}
\item
\begin{enumerate}
\item Montrer que: $\dfrac{7a+8c}{7b+8d}=\dfrac{a}{b}$
\item Montrer que: $\dfrac{a^2+b^2}{ab}=\dfrac{c^2+d^2}{cd}$
\item Montrer que: $\sqrt{\dfrac{a^2+c^2}{b^2+d^2}}=\dfrac{a}{b}$
\item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
\end{enumerate}
\item
\begin{enumerate}
\item Vérifier que si $\dfrac{a}{b}=\dfrac{c}{d}$ alors $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$.
\item Démontrer la réciproque, à savoir: si $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$ alors $\dfrac{a}{b}=\dfrac{c}{d}$
\\
\end{enumerate}
\end{enumerate}
\underline{\textbf{Exercice 4}}
\\
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels
\end{enumerate}
\end{tcolorbox}
\end{document}
Here is a screenshot of what I have. What I want to do is:
1) Making the exercises into twocolumn format.
2) Writing exercises title and numbering like in the first picture or any other styled template for exercises specially with two column.
Best Answer
like this?
With
enumitem
instead of theenumerate
package,enumerate
list styles are changed only intcolorbox
ses by help of `enumitem˙ package, considered @Leonardis comment:Edit:
exam
is right tool for you. For an example of it use see this answer.longtable
: