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Cel mai inteligent răspuns!
2014-05-17T23:47:20+03:00
Conditia de existenta: n natural, n≥8

\dfrac{n!}{(n-8)!}+\dfrac{n!}{(n-7)!}=9\cdot\dfrac{n!}{(n-6)!}

Tinem cont de faptul ca
(n-7)!=(n-8)!(n-7)\ si\ (n-6)!=(n-8)!(n-7)(n-6), dupa ce inmultim ecuatia cu \dfrac{(n-8)!}{n!}, obtinem:

1+\dfrac{1}{n-7}=\dfrac{9}{(n-7)(n-6)}, iar dupa aducerea la acelasi numitor, obtinem:

n^2-13n+42+n-6=9

n^2-12n+27=0

(n-3)(n-9)=0, care are solutiile 3 si 9, dar tinind cont de conditiile de existenta, singura solutie este n=9.


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