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

Răspuns :

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

[tex] a^{5} + 3 a^{4} \frac{1}{ a^{2} } +3 a^{2} \frac{1}{ a^{4} } + \frac{1}{ a^{5} } =2744[/tex] am simplificat a^4 cu a² 

[tex] a^{5} + 3 a^{2} +3 \frac{1}{ a^{2} } + \frac{1}{ a^{5} } =2744[/tex]

[tex]a^{5} + 3 ( a^{2} +\frac{1}{ a^{2} })+ \frac{1}{ a^{5} } =2744 [/tex]

[tex] a^{5} + 3*14+ \frac{1}{ a^{5} } =2744[/tex]

[tex] a^{5} + 42+ \frac{1}{ a^{5} } =2744[/tex]

[tex]a^{5} + \frac{1}{ a^{5} } =2744-42[/tex]

[tex]a^{5}+ \frac{1}{ a^{5} } =2702[/tex]