enter image description here

I want to make a result like this one, but I have no clue. Please give me your suggest. Thank you a lot for your help.

P/s: Please tell me how to typeset equations in the same way like this... Sorry my bad English cannot give you all what I'm asking for.

For specific:

\documentclass[a4paper , 12pt]{article}
\usepackage{color}
\usepackage[dvipsnames]{xcolor}
\usepackage{array}
\usepackage{floatrow}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}
\usepackage[utf8]{vietnam}
\usepackage{tikz,tkz-tab}
\usepackage{amssymb}
\usepackage{dsfont}
\usepackage{pgf,tikz,pgfplots}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usepackage{mathtools}
\usetikzlibrary{arrows}
\renewcommand{\baselinestretch}{1.5} %Chỉnh dãn dòng
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\makeatletter
\newcases{bcases}
  {\quad}
  {$\m@th{##}$\hfil}
  {$\m@th{##}$\hfil}
  {\lbrack}
  {.}
\makeatother

\begin{document}
{\color{blue}\textbf{Câu 18: }} Cho hàm số $y=f(x)$ có bảng biến thiên như hình phía dưới. Tìm tất cả $m\in \mathds{R}$ để hàm số $y=\dfrac{1}{f(x)-m}$ có 2 đường tiệm cận đứng. \\
\begin{center}
\begin{tikzpicture}
    \tkzTabInit{$x$ /1, $f'(x)$ /1, $f(x)$ /2.5} 
    {$-\infty$,-1,1,2, $+\infty$}
    \tkzTabLine{,+,0,-,0,+,0,-  }
    \tkzTabVar{-/$-\infty$, +/$4$ , -/$-5$, +/$4$, -/$\infty$}
\end{tikzpicture}
\end{center} 
\begin{tabular}{m{6cm}m{6cm}}
{\color{blue}\textbf{A.}} $m=4 v m<-5$ & {\color{blue}\textbf{B.}} $m=4$ \\
{\color{blue}\textbf{C.}} $m<-5$ & {\color{red}\textbf{D.}} \[
\begin{bcases}
m = A \\
m < -5
\end{bcases}
\] \\
\end{tabular}
\end{document}

enter image description here

The final picture is what I expected for

  • You can use $\begin{bcases}...\end{bcases}$ instead of \[...\], – egreg Sep 19 at 9:46
  • I added something based on your test case – egreg Sep 19 at 9:52
  • please tell us, what you expect from your code example? it works fine, but result is somehow unusual (to my taste) – Zarko Sep 19 at 9:54

The \newcases macro of mathtools ought to have an easier interface; with the current one you can still define a bcases environment (b for “bracket”).

\documentclass{article}
\usepackage{mathtools}

\makeatletter
\newcases{bcases}
  {\quad}
  {$\m@th{##}$\hfil}
  {$\m@th{##}$\hfil}
  {\lbrack}
  {.}
\makeatother

\begin{document}

\[
\begin{bcases}
m = A \\
m < -5
\end{bcases}
\]

\end{document}

enter image description here

You should also consider the more standard cases environment (it uses a brace).


Your test case can be improved:

\documentclass[a4paper , 12pt]{article}
\usepackage{color}
\usepackage[dvipsnames]{xcolor}
\usepackage{array}
\usepackage{floatrow}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}
\usepackage[utf8]{vietnam}
\usepackage{tikz,tkz-tab}
\usepackage{amssymb}
\usepackage{dsfont}
\usepackage{pgf,tikz,pgfplots}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usepackage{mathtools}
\usetikzlibrary{arrows}
\renewcommand{\baselinestretch}{1.5} %Chỉnh dãn dòng
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\makeatletter
\newcases{bcases}
  {\quad}
  {$\m@th{##}$\hfil}
  {$\m@th{##}$\hfil}
  {\lbrack}
  {.}
\makeatother

\begin{document}
\textcolor{blue}{\textbf{Câu 18: }} Cho hàm số $y=f(x)$ có bảng biến thiên 
như hình phía dưới. Tìm tất cả $m\in \mathds{R}$ để hàm số 
$y=\dfrac{1}{f(x)-m}$ có 2 đường tiệm cận đứng.
\begin{center}
\begin{tikzpicture}
    \tkzTabInit{$x$ /1, $f'(x)$ /1, $f(x)$ /2.5} 
    {$-\infty$,-1,1,2, $+\infty$}
    \tkzTabLine{,+,0,-,0,+,0,-  }
    \tkzTabVar{-/$-\infty$, +/$4$ , -/$-5$, +/$4$, -/$\infty$}
\end{tikzpicture}
\end{center} 
\begin{tabular}{p{6cm}p{6cm}}
\textcolor{blue}{\textbf{A.}} $m=4 \vee m<-5$ &
\textcolor{blue}{\textbf{B.}} $m=4$ \\
\textcolor{blue}{\textbf{C.}} $m<-5$ & \textcolor{red}{\textbf{D.}}
$\begin{bcases}
m = A \\
m < -5
\end{bcases}$
\end{tabular}
\end{document}

Note \textcolor{<color>}{<text>} instead of {\color{<color>}<text>}; the removal of useless \\ tokens; the p column type instead of m; $...$ around bcases instead of \[...\].

enter image description here

I use an aligned stack here only to emphasize that if the left hand side of the two rows had different widths (here they do not), the aligned forms would be the most appropriate.

\documentclass{article}
\usepackage{tabstackengine}
\stackMath
\begin{document}
\[
  \left[\alignCenterstack{m=&4\\m<&-5}\right.
\]
\end{document}

enter image description here

In terms of the OP's test case, I give two different vertical alignment options, in the form of "D" and "E".

\documentclass[a4paper , 12pt]{article}
\usepackage{color}
\usepackage[dvipsnames]{xcolor}
\usepackage{array}
\usepackage{floatrow}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}
\usepackage[utf8]{vietnam}
\usepackage{tikz,tkz-tab}
\usepackage{amssymb}
\usepackage{dsfont}
\usepackage{pgf,tikz,pgfplots}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usepackage{mathtools}
\usetikzlibrary{arrows}
\renewcommand{\baselinestretch}{1.5} %Chỉnh dãn dòng
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{tabstackengine}
\stackMath
\begin{document}
{\color{blue}\textbf{Câu 18: }} Cho hàm số $y=f(x)$ có bảng biến thiên như hình phía dưới. Tìm tất cả $m\in \mathds{R}$ để hàm số $y=\dfrac{1}{f(x)-m}$ có 2 đường tiệm cận đứng. \\
\begin{center}
\begin{tikzpicture}
    \tkzTabInit{$x$ /1, $f'(x)$ /1, $f(x)$ /2.5} 
    {$-\infty$,-1,1,2, $+\infty$}
    \tkzTabLine{,+,0,-,0,+,0,-  }
    \tkzTabVar{-/$-\infty$, +/$4$ , -/$-5$, +/$4$, -/$\infty$}
\end{tikzpicture}
\end{center} 
\begin{tabular}{m{6cm}m{6cm}}
{\color{blue}\textbf{A.}} $m=4 v m<-5$ & {\color{blue}\textbf{B.}} $m=4$ \\
{\color{blue}\textbf{C.}} $m<-5$ & {\color{red}\textbf{D.}} 
$\left[\alignCenterstack{m =& A \\m <& -5}\right.$ \\
{\color{red}\textbf{E.}} 
\belowbaseline[-\ht\strutbox]{\left[\alignCenterstack{m =& A \\m <& -5}\right.}
&
\end{tabular}
\end{document}

enter image description here

  • It doesn't work much in my test question :( Check the code of my test question please. – Hoàng Duy Anh Sep 19 at 9:51
  • @HoàngDuyAnh Yes it does. Please see my edit. – Steven B. Segletes Sep 19 at 9:59
  • Oh I see, that's exactly what I'm trying to text out! Thank you alot – Hoàng Duy Anh Sep 19 at 10:00

enter image description here

a simple way which cross my mind :-) :

\documentclass{article}

\begin{document}
\[
\left[
    \begin{array}{l}
        m=4 \\
        m<-5
    \end{array}
\right.
\]
\end{document}

edit: considering your mwe, with my above proposition you will obtain:

enter image description here

\documentclass[a4paper , 12pt]{article}
% \usepackage{color}  % it is loaded by xcolor
\usepackage[dvipsnames]{xcolor}
\usepackage{array}
\usepackage{floatrow}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}
\usepackage[utf8]{vietnam}
\usepackage{tikz,tkz-tab}
\usepackage{amssymb}
\usepackage{dsfont}
\usepackage{pgfplots}%pgf,tikz
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usepackage{mathtools}
\usetikzlibrary{arrows}
\renewcommand{\baselinestretch}{1.5} %Chỉnh dãn dòng
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}


\begin{document}
\textcolor{blue}{\textbf{Câu 18: }} Cho hàm số $y=f(x)$ có bảng biến thiên
như hình phía dưới. Tìm tất cả $m\in \mathds{R}$ để hàm số
$y=\dfrac{1}{f(x)-m}$ có 2 đường tiệm cận đứng.
\begin{center}
\begin{tikzpicture}
    \tkzTabInit{$x$ /1, $f'(x)$ /1, $f(x)$ /2.5}
    {$-\infty$,-1,1,2, $+\infty$}
    \tkzTabLine{,+,0,-,0,+,0,-  }
    \tkzTabVar{-/$-\infty$, +/$4$ , -/$-5$, +/$4$, -/$\infty$}
\end{tikzpicture}
\end{center}
\begin{tabular}{p{6cm}p{6cm}}
\textcolor{blue}{\textbf{A.}} $m=4 \vee m<-5$ &
\textcolor{blue}{\textbf{B.}} $m=4$ \\
\textcolor{blue}{\textbf{C.}} $m<-5$ & \textcolor{red}{\textbf{D.}}
            $\left[\begin{array}{c}
                        m = A \\
                        m < -5
                        \end{array}\right.$
\end{tabular}
\end{document}

\documentclass[t]{beamer}
\setbeamertemplate{navigation symbols}{}
\setbeamertemplate{footline}{}
\setbeamertemplate{headline}{}
\usepackage[edges]{forest}
  • Oh thank you a lots, That's help a lot but there is some struggle when I put it in the answers of my test: – Hoàng Duy Anh Sep 19 at 9:15
  • 2
    @HoàngDuyAnh - Please be more specific about the "struggle when I put in my answers" consists of. – Mico Sep 19 at 9:40
  • @HoàngDuyAnh, you not provide any code (as small complete document which we can copy and test), so we haven't any idea about your document. anyway, the proposed solution is so elementary that it should work in any document. – Zarko Sep 19 at 9:40
  • Let me add my code into the question – Hoàng Duy Anh Sep 19 at 9:40

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