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A melhor resposta!
2014-07-15T15:48:36-03:00
Olá Duda,

para facilitar, multiplique cruzado e faça as somas, multiplicando os denominadores também:

 \dfrac{5}{x}+ \dfrac{4}{x+1}= \dfrac{7}{12}~\to~ \dfrac{5*(x+1)+4*x}{x*(x+1)}= \dfrac{7}{12}\\\\\\
~\to~ \dfrac{5x+5+4x}{ x^{2} +x}= \dfrac{7}{12}~\to~ \dfrac{9x+5}{ x^{2} +x}= \dfrac{7}{12}\\\\\\
~\to~7( x^{2} +x)=12(9x+5)~\to~7 x^{2} +7x=108x+60\\\\
~\to~7 x^{2} +7x-108x-60=0~\to~7 x^{2} -101x-60=0

\Delta=b^2-4ac\\
\Delta=(-101)^2-4*7*(-60)\\
\Delta=10.201+1.680\\
\Delta=11.881

x= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-101)\pm \sqrt{11.881} }{2*7}= \dfrac{101\pm109}{14}\\\\\\
~\to~\begin{cases}x'= \dfrac{101+109}{14}= \dfrac{210}{14} =15\\\\
x''= \dfrac{101-109}{14}= \dfrac{-8}{~14}= \dfrac{-8:2}{~14:2}=- \dfrac{4}{7}\end{cases}\\\\\\
S=\left\{15,- \dfrac{4}{7}\right\}

Espero ter ajudado e tenha ótimos estudos =))


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