Difference between revisions of "LaTeX"

Latest revision as of 18:19, 11 May 2022

LaTeX
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The $\text{\LaTeX}$ typesetting system (typically pronounced "Lah-Tek", but also sometimes "Lay-tek", "Lah-tekh", or "Lah-tex") is widely used to produce well-formatted mathematical and scientific writing. $\text{\LaTeX}$ is very handy for producing equations such as

$1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.$

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

Click here to start learning how to use $\text{\LaTeX}$ on AoPS.
Click here for a Quick Start $\text{\LaTeX}$ video.
Click here for a guide to making .pdfs using $\text{\LaTeX}$ for use unrelated to AoPS.
Click here for a wonderful guide for more professional $\text{\LaTeX}$ by Overleaf.

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-<math>5</math>)Resolva em U=R, onde a incógnita é x e <math>a\neq3</math>
+{{Latex}}
+The <math>\text{\LaTeX}</math> typesetting system (typically pronounced "Lah-Tek", but also sometimes "Lay-tek", "Lah-tekh", or "Lah-tex") is widely used to produce well-formatted [[math|mathematical]] and scientific writing. <math>\text{\LaTeX}</math> is very handy for producing equations such as
-<math>a(x+1)=3x-2</math>
+<cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>
-Resp:
+Nearly every serious student of math or science will use <math>\text{\LaTeX}</math> frequently. Through these web pages, you will learn much of what you'll need to express math and science like a professional, much, much better than writing papers with Word!
-<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>
+* Click [http://www.artofproblemsolving.com/wiki/index.php/LaTeX:LaTeX_on_AoPS here] to start learning how to use <math>\text{\LaTeX}</math> on AoPS.
+* Click [https://youtu.be/HfRsDKBrSaA here] for a Quick Start <math>\text{\LaTeX}</math>  video.
-<math>8</math>) Resolva a equação:
+* Click [http://latex-tutorial.com/ here] for a guide to making .pdfs using <math>\text{\LaTeX}</math> for use unrelated to AoPS.
+* Click [https://www.overleaf.com/learn here] for a wonderful guide for more professional <math>\text{\LaTeX}</math> by Overleaf.
-<math>\bullet</math> No papel que recebi estava assim:
-<math>\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{2}+\dfrac{1}{x-2}</math> em U=R-{2,0}
-<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.

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Difference between revisions of "LaTeX"

Latest revision as of 18:19, 11 May 2022