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Difference between revisions of "LaTeX"
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<math>5</math>)Resolva em U=R, onde a incógnita é x e <math>a\neq3</math>  
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{{Latex}}
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The <math>\LaTeX</math> typesetting system (pronounced "Lay-Tek" by most, or "Lah-Tek" by some) is widely used to produce well-formatted [[math|mathematical]] and scientific writing. With <math>\LaTeX</math>, it is very easy to produce expressions like
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<cmath>
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\sqrt{\frac {a^2+b^2+c^2}3} \geq \frac {a+b+c}3 \geq \sqrt[3]{abc} \geq \frac 3 { \frac 1a + \frac 1b + \frac 1c } .
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</cmath> Nearly every serious student of math and science will use <math>\LaTeX</math> frequently. Through these web pages, you will learn much of what you'll need to express math and science like a pro.
  
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<math>a(x+1)=3x-2</math>
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* [http://www.artofproblemsolving.com/wiki/index.php/LaTeX:LaTeX_on_AoPS Click here] to start learning how to use <math>\LaTeX</math> on AoPS!
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Resp:
 
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<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>
 
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<math>8</math>) Resolva a equação:
 
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<math>\bullet</math> No papel que recebi estava assim:
 
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<math>\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{2}+\dfrac{1}{x-2}</math> em U=R-{2,0}
 
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<math>\bullet</math> acho que esse sinal de menos no meio era pra ser um igual.
 
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<math>\dfrac{1}{x-2}+\dfrac{1}{x}=\dfrac{1}{2}+\dfrac{1}{x-2}</math>
 
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<math>\bullet</math> porém ainda fica estranho porque a solução seria <math>x=2</math>.
 
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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.
 
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<math>9</math>) Resolva a equação
 
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<math>\dfrac{1}{x-3}+\dfrac{1}{x}=\dfrac{1}{3}+\dfrac{1}{x-3}</math> em U=R-{3,0}
 
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<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.
 
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Portanto a questão não tem solução.
 
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<math>10</math>) Resolva as equações:
 
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b) <math>\dfrac{1}{a}+\dfrac{a}{a+x}=\dfrac{a+x}{ax}</math>
 
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<math>\bullet</math> Ele deveria ter mencionado quem era incógnita e quem era constante. Pelas outras questões supomos que x seja a incógnita.
 
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<math>\bullet</math> Ele também não mencionou que o 'a' e o 'x' tem que ser diferente de 0.
 
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<math>\bullet</math> Considerando o que foi dito acima, vem:
 
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<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>
 
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<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>
 
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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>
 
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<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>
 
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Não tenho certeza se o que eu fiz está certo.
 

Revision as of 19:05, 28 November 2015

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

The $\LaTeX$ typesetting system (pronounced "Lay-Tek" by most, or "Lah-Tek" by some) is widely used to produce well-formatted mathematical and scientific writing. With $\LaTeX$, it is very easy to produce expressions like \[\sqrt{\frac {a^2+b^2+c^2}3} \geq \frac {a+b+c}3 \geq \sqrt[3]{abc} \geq \frac 3 { \frac 1a + \frac 1b + \frac 1c } .\] Nearly every serious student of math and science will use $\LaTeX$ frequently. Through these web pages, you will learn much of what you'll need to express math and science like a pro.

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