Pré-Vestibular ⇒ (PUCCAMP) Frações Algébricas Tópico resolvido

[bruninha]

Se [tex3]x[/tex3] e [tex3]y[/tex3] são números reais tais que [tex3]x.y \,\neq \, 0[/tex3] e [tex3]|x|\, \neq \, |y|[/tex3] a expressão [tex3]\frac { x^{-8}\, - \, y^{-8} } {x^{-2}\,.\,y^{-2}\,.\,(x^{-4}\, + \, y^{-4})}[/tex3]

a) [tex3]x^2y^2\,(x-y)(x+y)[/tex3]
b) [tex3]x^2y^2\,(x-y)^2[/tex3]
c) [tex3]\left(\frac{y}{x}\, - \, \frac{x}{y}\right)\,\cdot \,\left(\frac{y}{x}\, + \, \frac{x}{y}\right)[/tex3]
d) [tex3]xy\, (x^2\,+\,y^2)[/tex3]
e) [tex3]xy\, \left(\frac{y^2}{x^2}\, - \, \frac{x^2}{y^2}\right)[/tex3]

Resposta:

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c

[bigjohn]

• [tex3]\frac{x^{-8}-y^{-8}}{x^{-2}\cdot y^{-2}\cdot (x^{-4}+y^{-4})}= \frac{\frac{1}{x^{8}}-\frac{1}{y^{8}}}{\frac{1}{x^{2}}\cdot \frac{1}{y^{2}}\cdot\left(\frac{1}{x^{4}}+\frac{1}{y^{4}}\right)}=[/tex3]
  
  [tex3]\frac{\frac{y^8-x^8}{x^{8}\cdot y^8}}{\frac{1}{x^2\cdot y^2}\left(\frac{y^4+x^4}{x^4\cdot y^4}\right)}= \frac{\frac{y^8-x^8}{x^{8}\cdot y^8}}{\left(\frac{y^4+x^4}{x^6\cdot y^6}\right)}= \frac{y^8-x^8}{x^{8}\cdot y^8}\cdot\frac{x^6\cdot y^6}{y^4+x^4}= \frac{(y^8-x^8)\cdot(x^6\cdot y^6)}{(x^{8}\cdot y^8)\cdot(y^4+x^4)}=[/tex3]
  
  [tex3]\frac{(y^8-x^8)}{(x^2\cdot y^2)\cdot(y^4+x^4)}= \frac{(y^4-x^4)(y^4+x^4)}{(x^2\cdot y^2)\cdot(y^4+x^4)}= \frac{y^4-x^4}{x^2\cdot y^2}=\frac{y^4}{x^2\cdot y^2}-\frac{x^4}{x^2\cdot y^2}=\frac{y^2}{x^2}-\frac{x^2}{y^2}=[/tex3]
  
  [tex3]\left(\frac{y}{x}\right)^2-\left(\frac{x}{y}\right)^2 =\left(\frac{y}{x}-\frac{x}{y}\right)\cdot\left(\frac{y}{x}+\frac{x}{y}\right)[/tex3]

[fabit]

• [tex3]\frac{\(\frac{1}{x}\)^8-\(\frac{1}{y}\)^8}{\frac{1}{x^2y^2}\(\(\frac{1}{x}\)^4+\(\frac{1}{y}\)^4\)}=\frac{x^2y^2\cancel{\(\frac{1}{x^4}+\frac{1}{y^4}\)}\(\frac{1}{x^4}-\frac{1}{y^4}\)}{\cancel{\frac{1}{x^4}+\frac{1}{y^4}}}=\frac{y^2}{x^2}-\frac{x^2}{y^2}=\(\frac{y}{x}+\frac{x}{y}\)\(\frac{y}{x}-\frac{x}{y}\)[/tex3]
  
  ou [tex3]\frac{y^4-x^4}{x^2y^2}[/tex3]