24 fev 2013, 20:41
se x>0 e (x^2)+ 1/(2^2)=7
(x^5)+1/(x^5) = ?
24 fev 2013, 21:11
Presumo que seja \(x^2 + \frac{1}{x^2} = 7\).
Se sim, segue que:
\(x^2 + \frac{1}{x^2} = 7\)
\(\left ( x + \frac{1}{x} \right )^2 - 2 = 7\)
\(\left ( x + \frac{1}{x} \right )^2 = 9\)
\(\left ( x + \frac{1}{x} \right ) = \sqrt{9}\)
\(\fbox{\left ( x + \frac{1}{x} \right ) = 3}\)
elevando ao cubo...
\(x^3 + 3x^2 \cdot \frac{1}{x} + 3x \cdot \frac{1}{x^2} + \frac{1}{x^3} = 27\)
\(x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} = 27\)
\(x^3 + 3\left (x + \frac{1}{x} \right ) + \frac{1}{x^3} = 27\)
\(x^3 + 3 \cdot 3 + \frac{1}{x^3} = 27\)
\(\fbox{x^3 + \frac{1}{x^3} = 18}\)
Com efeito,
\(\left ( x^2 + \frac{1}{x^2} \right ) \cdot \left ( x^3 + \frac{1}{x^3} \right ) = 7 \cdot 18\)
\(x^5 + \frac{x^2}{x^3} + \frac{x^3}{x^2} + \frac{1}{x^5} = 126\)
\(x^5 + \frac{1}{x} + x + \frac{1}{x^5} = 126\)
\(x^5 + \left (x + \frac{1}{x} \right ) + \frac{1}{x^5} = 126\)
\(x^5 + 3 + \frac{1}{x^5} = 126\)
\(\fbox{\fbox{\fbox{x^5 + \frac{1}{x^5} = 123}}}\)