3

enter image description here

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.

enter image description here

8
  • 1
    Why TikZ and tcolorbox tags?
    – Ignasi
    Jun 7 '19 at 18:18
  • Could you please show what you have tried so far? Please also explicitly state your problem. Is it the chapter title, the colored boxes around the numbers or something completely different?
    – leandriis
    Jun 7 '19 at 18:22
  • Regarding the colored boxes you might be interested in Colored box around enumerate labels
    – leandriis
    Jun 7 '19 at 18:23
  • 1
    @HAMIDINESOUMARE: Your question is a legitimate one. Unfornunately, too much graphic diversity will not add to the clarity of the generated output.
    – Marian G.
    Jun 7 '19 at 18:48
  • 1
    Welcome to TeX.SX! Can you please expand the code snippet that you have posted to a full minimal working example. It is much easier to help you if we can start with some compilable code that illustrates your problem. A MWE should start with a \documentclass command, include any necessary packages and be as small as possible to demonstrate your problem. At the moment we have to guess what packages etc you are using before we can compile your code.
    – user30471
    Jun 8 '19 at 0:55
7

like this?

enter image description here enter image description here

With enumitem instead of the enumerate package, enumerate list styles are changed only in tcolorboxses by help of `enumitem˙ package, considered @Leonardis comment:

\documentclass[11pt]{book}
\usepackage[margin=2.5cm]{geometry}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}
\usepackage{mathrsfs}

\usepackage{amsmath,amssymb}
\usepackage[most]{tcolorbox}
\newtcolorbox{exercisebox}%
    {
    enhanced,
    breakable,
    colback=white, colframe=green!15!white,
    colbacktitle=white!15!pink, coltitle=pink!50!black,
    left=0pt,right=0mm,top=3mm,bottom=3mm,
    pad at break=0pt,bottomrule at break=0pt,toprule at break=0pt,
    borderline={0mm}{0mm}{green!50!white,dashed},
    attach boxed title to top center={yshift=-2mm},
    boxed title style={boxrule=0.4pt},
    title=EXERCICES,
    }

\usepackage{eso-pic}
\usepackage{etoolbox}
\usepackage{enumitem}
\newcommand\circitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[circle,draw=gray, fill=gray!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\newcommand\boxitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[fill=orange!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\AtBeginEnvironment{tcolorbox}{%
    \setlist[enumerate,1]{label=\protect\boxitem{\arabic*}}
    \setlist[enumerate,2]{label=\protect\circitem{\alph*}}
                          }
\usepackage{fancyhdr}
\usepackage{cancel}
\usepackage{xcolor}
\newcommand\Exercice[1]{\fboxrule1.5pt%
\fcolorbox{gray}% frame
          {cyan}% background
          {\large\itshape\textbf{Exercise~#1}}%
          \par\medskip}

\usepackage{graphicx}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{arrows}
\usepackage{tkz-tab}
\usepackage{mdframed}

\usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{hyperref}


\usepackage{lipsum}

\begin{document}
\section{EXERCICES}

\begin{exercisebox}
\Exercice{1}
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\mfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \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}

\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}

\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}

\Exercice{4}
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels
\begin{enumerate}
 \item Vérifier les identités suivantes:
\begin{enumerate}
\item $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
    \item $(ax+by)^2-(ay+bx)^2=(a^2-b^2)(x^2-y^2)$
    \item $(a^2+a+1)(a^2-a+1)=a^4+a^2+1$
    \end{enumerate}
    \item
    \begin{enumerate}
    \item Montrer l'identité de Gauss à savoir: \\
    $a^3+b^2+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
        \item Servez vous de cette relation pour montrer que pour tout $u,v$ et $w$ on a l'identité:
    \\ $(u-v)^3+(v-w)^3+(w-u)^3=3(u-v)(v-w)(w-u)$
    \end{enumerate}
\end{enumerate}

\Exercice{5}
Factoriser les expressions suivantes
\begin{enumerate}
    \item $ab+a+b+1$
    \item $a^2xy+aby^2+b^2xy+abx^2$
    \item $a^2+8+2a^2+4a$
    \item $ab(a+b)+bc(b+c)+ca(c+a)+2abc$
    \item $(ax+by)^2-(ay+bx)^2$
    \item $(a^2+b^2-9)^2-4a^2b^2$
    \item $(a+b)^2-a^3-b^3$
    \item $(a^4-b^4)+2ab(a^2-b^2)-(a^3-b^2)+(ab^2-a^2b)$
    \item $x^2y - xy^2 + yz^2 - xz^2 + x^2z - xyz + y^2z + xyz$
\end{enumerate}
\end{exercisebox}
\end{document}

Edit:

  • If you consider less fancy design of your exercises, than the package exam is right tool for you. For an example of it use see this answer.
  • With use of above proposed solution for questions labels is possible to arrange them in two column by manula insert each question in appropriate place in longtable:

\documentclass[11pt]{book}
\usepackage[margin=2.5cm]{geometry}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}
\usepackage{mathrsfs}

\usepackage{nccmath,amssymb}
\usepackage[most]{tcolorbox}
\newtcolorbox{exercisebox}%
    {
    enhanced,
    breakable,
    colback=white, colframe=green!15!white,
    colbacktitle=white!15!pink, coltitle=pink!50!black,
    left=0pt,right=0mm,top=3mm,bottom=3mm,
    pad at break=0pt,bottomrule at break=0pt,toprule at break=0pt,
    borderline={0mm}{0mm}{green!50!white,dashed},
    attach boxed title to top center={yshift=-2mm},
    boxed title style={boxrule=0.4pt},
    title=EXERCICES,
    }

\usepackage{eso-pic}
\usepackage{enumitem}
\newcommand\circitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[circle,draw=gray, fill=gray!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}
\newcommand\boxitem[1]{%
    \tikz[baseline=(char.base)]{%https://tex.stackexchange.com/questions/204116/uniform-size-of-circles-around-enumitems
        \node[fill=orange!30,
              minimum size=1.2em,inner sep=0] (char) {#1};}}

\usepackage{fancyhdr}
\usepackage{cancel}
\usepackage{xcolor}
\newcommand\Exercice[1]{\fboxrule1.5pt%
\fcolorbox{gray}% frame
          {cyan}% background
          {\large\itshape\textbf{Exercise~#1}}%
          \par\medskip}

\usepackage{graphicx}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{arrows}
\usepackage{tkz-tab}
\usepackage{mdframed}

\usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{multicol}
\usepackage{longtable}
\usepackage{hyperref}



\usepackage{lipsum}

\begin{document}
\lipsum[66]
\setcounter{chapter}{7}
\section{EXERCICES}

\begin{exercisebox}
    \setlist[enumerate,1]{label=\protect\boxitem{\arabic*},leftmargin=*,resume}
    \setlist[enumerate,2]{label=\protect\circitem{\alph*}}
\begin{longtable}{@{}*{2}{p{\dimexpr0.5\linewidth-\tabcolsep}}}

\Exercice{1}
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\mfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \item $B=\mfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
    \item $C=\mfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
    \item $D=\mfrac{\left[\left(\mfrac23\right)^2\right]^6\times \left[\left(\mfrac35\right)^{-2}\right]^3\times\left[\left(\mfrac52\right)^2\right]^{-6}}{\left(\mfrac46\right)^6}$
\end{enumerate}
&
\Exercice{3}
\begin{enumerate}
    \item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.

Montrer que $\mfrac{b+c}{a}+\mfrac{c+a}{b}+\mfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:

$\mfrac{a}{ab+a+1}+\mfrac{b}{bc+c+1}+\mfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:

$\mfrac{4a^2-1}{(a-b)(a-c)}+\mfrac{4b^2-1}{(b-c)(b-a)}+\mfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}
\\
% new row
\Exercice{2}
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\mfrac{a}{b}=\mfrac{c}{d}$.
\begin{enumerate}
    \item
    \begin{enumerate}
        \item Montrer que: $\mfrac{7a+8c}{7b+8d}=\mfrac{a}{b}$
        \item Montrer que: $\mfrac{a^2+b^2}{ab}=\mfrac{c^2+d^2}{cd}$
        \item Montrer que: $\sqrt{\mfrac{a^2+c^2}{b^2+d^2}}=\mfrac{a}{b}$
        \item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
    \end{enumerate}
    \item
    \begin{enumerate}
        \item Vérifier que si $\mfrac{a}{b}=\mfrac{c}{d}$ alors $\mfrac{a^2+b^2}{ac+bd}=\mfrac{ac+bd}{c^2+d^2}$.
        \item Démontrer la réciproque, à savoir: si $\mfrac{a^2+b^2}{ac+bd}=\mfrac{ac+bd}{c^2+d^2}$ alors $\mfrac{a}{b}=\mfrac{c}{d}$
    \end{enumerate}
\end{enumerate}
&
\Exercice{4}
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels
\begin{enumerate}
 \item Vérifier les identités suivantes:
    \begin{enumerate}
    \item $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
    \item $(ax+by)^2-(ay+bx)^2=(a^2-b^2)(x^2-y^2)$
    \item $(a^2+a+1)(a^2-a+1)=a^4+a^2+1$
    \end{enumerate}
\item
    \begin{enumerate}
    \item Montrer l'identité de Gauss à savoir:

    $a^3+b^2+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
        \item Servez vous de cette relation pour montrer que pour tout $u,v$ et $w$ on a l'identité:

    $(u-v)^3+(v-w)^3+(w-u)^3=3(u-v)(v-w)(w-u)$
    \end{enumerate}
\end{enumerate}
\\
% new row
\Exercice{5}
Factoriser les expressions suivantes
\begin{enumerate}
    \item $ab+a+b+1$
    \item $a^2xy+aby^2+b^2xy+abx^2$
    \item $a^2+8+2a^2+4a$
    \item $ab(a+b)+bc(b+c)+ca(c+a)+2abc$
\end{enumerate}
&
\begin{enumerate}[start=5]
    \item $(ax+by)^2-(ay+bx)^2$
    \item $(a^2+b^2-9)^2-4a^2b^2$
    \item $(a+b)^2-a^3-b^3$
    \item $(a^4-b^4)+2ab(a^2-b^2)-(a^3-b^2)+(ab^2-a^2b)$
    \item $x^2y - xy^2 + yz^2 - xz^2 + x^2z - xyz + y^2z + xyz$
\end{enumerate}
\end{longtable}
\vspace{-2\baselineskip}
\end{exercisebox}
\lipsum[3]
\end{document}

enter image description here

enter image description here

7
  • thank you a lot. This is so useful to me. I have a question. Is there a why to split it into two columns.
    – DINEDINE
    Jun 8 '19 at 5:07
  • @HAMIDINESOUMARE, I do not understand, what you mean with "two columns#"? That (only) this section is in two columns or whole document?
    – Zarko
    Jun 8 '19 at 6:59
  • Only exercises section.
    – DINEDINE
    Jun 8 '19 at 7:03
  • @HAMIDINESOUMARE, see edited answer. Is this what you looking for?
    – Zarko
    Jun 8 '19 at 15:18
  • Not exactly. This is what I had earlier. What I am looking for is just a line separating the box into two column not splitting the previous box into two.
    – DINEDINE
    Jun 8 '19 at 15:23
4

The only difference between the approach below and Zarko's nice answer, or even the MWE, is largely presentation. I like having environments that take care of "all" of the presentation and I like names that "mean something". So, I would wrap everything into a new exercises environment that allows you to type:

\begin{exercises}
   \exercise first exercise
      \begin{questions}
        \question first question
        \question second question
        ...
      \end{questions}
   \exercise second exercise
   ...
\end{exercises}

in order to create everything. Under the hood these would both be enumerate environments, defined using the enumitem package, and I would use tikz for the fancy labels (following the recipes from Colored box around enumerate labels) and Forcing new line after item number in enumerate environment.

Here's the output from the code below:

enter image description here

and here is the code:

\documentclass[11pt]{book}
\usepackage[top=1cm, bottom=1cm, left=1.5cm, right=1.5cm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[most]{tcolorbox}
\usepackage{eso-pic}
\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}

\newcommand{\fancylabel}[2]{\tikz[baseline=(current bounding box.center)]\node[#1]{#2};}
\usepackage{enumitem}
\newlist{exercises}{enumerate}{1}
\let\realitem\item
\newcommand\exercise[1][]{\realitem\mbox{}#1\\}% to for ce alne break after the item
\let\question\realitem
\setlist[exercises]{
  label=\protect\fancylabel{rectangle,fill=blue!20}{Exercise~\arabic*},
  before=\let\item\exercise,
  align=left,
}

\tcolorboxenvironment{exercises}{
  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,
}
\newlist{questions}{enumerate}{1}
\setlist[questions]{
  label=\protect\fancylabel{circle,fill=brown!20}{\arabic*},
  before=\let\item\question,
  leftmargin=*
}
\begin{document}

\begin{exercises}

  \exercise Simplifier les expressions suivantes:
  \begin{questions}
      \question $A=\dfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
      \question $B=\dfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
      \question $C=\dfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
      \question $D=\dfrac{\left[\left(\dfrac23\right)^2\right]^6\times \left[\left(\dfrac35\right)^{-2}\right]^3\times\left[\left(\dfrac52\right)^2\right]^{-6}}{\left(\dfrac46\right)^6}$
  \end{questions}

  \exercise
  \begin{questions}
      \question Soient $a,b$ et $c$ trois nombres réels non nuls tels que
      $ab+bc+ca=0$.\\ Montrer que
      $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=-3$
      \question On suppose maintenant que $abc=1$. Montrer que: \\
      $\dfrac{a}{ab+a+1}+\dfrac{b}{bc+c+1}+\dfrac{c}{ca+a+1}=1$
      \question 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{questions}

  \exercise Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\dfrac{a}{b}=\dfrac{c}{d}$.
  \begin{questions}
      \question Montrer que: $\dfrac{7a+8c}{7b+8d}=\dfrac{a}{b}$
      \question Montrer que: $\dfrac{a^2+b^2}{ab}=\dfrac{c^2+d^2}{cd}$
      \question Montrer que: $\sqrt{\dfrac{a^2+c^2}{b^2+d^2}}=\dfrac{a}{b}$
      \question Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
  \end{questions}

  \exercise
  \begin{questions}
      \question Vérifier que si $\dfrac{a}{b}=\dfrac{c}{d}$ alors $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$.
      \question 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{questions}

\exercise[Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels]

\end{exercises}
\end{document}

Rather than using \exercise and \question you can, in fact, use just \item but I prefer the "named" versions. As the last exercise shows, the \exercise accepts an optional argument that puts text on the same line as the exercise number.

1
  • Thank you for your helpful answer
    – DINEDINE
    Jun 8 '19 at 6:51

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