Question

Cum se face? Ca nu știu.

Answer 1

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Answer 2

n! /12=(n-2)! a.i. n∈N si n >=2 .

(n-2)!·(n-1)·n /12=(n-2)! <=> (n-2)!·(n+1)·n =12·(n-2)! dar deoarece n∈N a.i. n >=2 => n·(n+1)=12 si n <n+1 unde n·(n+1)=12=3·4 => n=3 ,dar observam ca 3! /12=1! <=> 6/12=1 <=> 1/2=1 ,contradictie => n=n+1=4 ,care verifica egalitatea deoarece 4! /12=2! <=> 24/12=2 <=> 2=2 ,care este adevarat .