Question

Fie a > 0,a E numerelor reale astfel incat a patrat +1supra a patrat =14.Calculati a la a5a +1 supra a la a5a .

Answer 1

a²+$\frac{1}{ a^{2} }$ =14 ridicam la a3 si vom avea inainte de egal formula (a+b)³ si dupa egal 14³

$a^{5} + 3 a^{4} \frac{1}{ a^{2} } +3 a^{2} \frac{1}{ a^{4} } + \frac{1}{ a^{5} } =2744$ am simplificat a^4 cu a² 

$a^{5} + 3 a^{2} +3 \frac{1}{ a^{2} } + \frac{1}{ a^{5} } =2744$

$a^{5} + 3 ( a^{2} +\frac{1}{ a^{2} })+ \frac{1}{ a^{5} } =2744$

$a^{5} + 3*14+ \frac{1}{ a^{5} } =2744$

$a^{5} + 42+ \frac{1}{ a^{5} } =2744$

$a^{5} + \frac{1}{ a^{5} } =2744-42$

$a^{5}+ \frac{1}{ a^{5} } =2702$