Difference between revisions of "LaTeX"

Latest revision as of 18:51, 10 March 2025

LaTeX
About - Getting Started - Diagrams - Symbols - Downloads - Basics - Math - Examples - Pictures - Layout - Commands - Packages - Help

The $\textbf{\LaTeX}$ typesetting markup language (typically pronounced "lah-tek", but also sometimes "lay-tek", or "lay-tex") is widely used to produce well-formatted mathematical and scientific writing. $\text{\LaTeX}$ is very handy for writing equations such as

$\sqrt[3]{\frac{4^2}{5}+\pi} \approx 1.9$

Nearly every serious student of math or science will use $\text{\LaTeX}$ frequently. Through these websites, you will learn what you'll need to express maths and science like a professional, much, much better than Microsoft Word!

Click here to start learning how to use $\text{\LaTeX}$ on AoPS.
LaTeX and Asymptote forum on AoPS
Click here for a guide for more professional $\text{\LaTeX}$ by Overleaf.
Click here for a guide to making PDFs using $\text{\LaTeX}$ for use unrelated to AoPS.
See or this 5-minute video or this 15-minute crash course on $\text{\LaTeX}$ .

@@ Line 1: / Line 1: @@
-<math>5</math>)Resolva em U=R, onde a incógnita é x e <math>a\neq3</math>
+{{latex}}
-<math>a(x+1)=3x-2</math>
+The <math>\textbf{\LaTeX}</math> typesetting markup language (typically pronounced "lah-tek", but also sometimes "lay-tek", or "lay-tex") is widely used to produce well-formatted [[math]]ematical and [[scientific]] writing. <math>\text{\LaTeX}</math> is very handy for writing equations such as
-Resp:
+<cmath>\sqrt[3]{\frac{4^2}{5}+\pi} \approx 1.9</cmath>
-<math>ax+a=3x-2 \to ax-3x=-(2+a) \to x=\dfrac{-(2+a)}{a-3} \to x=\dfrac{(2+a)}{3-a}</math>
+Nearly every serious student of math or science will use <math>\text{\LaTeX}</math> frequently. Through these websites, you will learn what you'll need to express maths and science like a professional, much, much better than Microsoft Word!
-<math>8</math>) Resolva a equação:
+* Click [[LaTeX:LaTeX_on_AoPS|here]] to start learning how to use <math>\text{\LaTeX}</math> on AoPS.
+* [{{SERVER}}/community/c68 LaTeX and Asymptote forum on AoPS]
+* Click [//www.overleaf.com/learn here] for a guide for more professional <math>\text{\LaTeX}</math> by Overleaf.
+* Click [//latex-tutorial.com here] for a guide to making PDFs using <math>\text{\LaTeX}</math> for use unrelated to AoPS.
+* See  or [//youtu.be/HfRsDKBrSaA this 5-minute video] or [//youtu.be/mdFD9uQNU24 this 15-minute crash course] on <math>\text{\LaTeX}</math>.
-<math>\bullet</math> No papel que recebi estava assim:
+== See Also ==
-<math>\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{2}+\dfrac{1}{x-2}</math> em U=R-{2,0}
+* [[Asymptote (Vector Graphics Language)]]
-<math>\bullet</math> acho que esse sinal de menos no meio era pra ser um igual.
-<math>\dfrac{1}{x-2}+\dfrac{1}{x}=\dfrac{1}{2}+\dfrac{1}{x-2}</math>
-<math>\bullet</math> porém ainda fica estranho porque a solução seria <math>x=2</math>.
-Não pode ser <math>x=2</math> porque a questão pede todas as soluções menos o 2 e o 0.Assim acredito que a questão não tenha solução.
-<math>9</math>) Resolva a equação
-<math>\dfrac{1}{x-3}+\dfrac{1}{x}=\dfrac{1}{3}+\dfrac{1}{x-3}</math> em U=R-{3,0}
-<math>\bullet</math> Da mesma forma que ocorreu na questão anterior a solução seria <math>x=3</math>, porém a questão pede todas as soluções menos o 3 e o 0.
-Portanto a questão não tem solução.
-<math>10</math>) Resolva as equações:
-b) <math>\dfrac{1}{a}+\dfrac{a}{a+x}=\dfrac{a+x}{ax}</math>
-<math>\bullet</math> Ele deveria ter mencionado quem era incógnita e quem era constante. Pelas outras questões supomos que x seja a incógnita.
-<math>\bullet</math> Ele também não mencionou que o 'a' e o 'x' tem que ser diferente de 0.
-<math>\bullet</math> Considerando o que foi dito acima, vem:
-<math>\dfrac{1}{a}+\dfrac{a}{a+x}=\dfrac{a+x}{ax}</math> <math>\to</math> <math>\dfrac{a+x+a^2}{a(a+x)}=\dfrac{a+x}{ax}</math> <math>\to</math> <math>xa+x^2+a^2x=(a+x)^2</math> <math>\to</math>
-<math>xa+x^2+a^2x=a^2+x^2+2ax</math> <math>\to</math> <math>a^2x+ax-2ax-a^2=0</math> <math>\to</math> <math>a^2x-ax-a^2=0</math> <math>\to</math> <math>(a^2-a)x=a^2</math> <math>\to</math> <math>x=\dfrac{a^2}{a^2-a}</math> <math>\to</math> <math>x=\dfrac{a}{a-1}</math>
-d) <math>\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{8}{x-1}</math> <math>\to</math> <math>\dfrac{x+1}{x-1}-\dfrac{8}{x-1}=\dfrac{x-1}{x+1}</math> <math>\to</math> <math>\dfrac{x-7}{x-1}=\dfrac{x-1}{x+1}</math> <math>\to</math>
-<math>(x-7)(x+1)=(x-1)^2</math> <math>\to</math> <math>x^2+x-7x-7=x^2+1-2x</math> <math>\to</math> <math>-6x-7=1-2x</math> <math>\to</math> <math>x=-2</math>
-Não tenho certeza se o que eu fiz está certo.

Page

Toolbox

Search

Difference between revisions of "LaTeX"

Latest revision as of 18:51, 10 March 2025

See Also