Bonjour
Domaine : n - 2 ≥ 0 (√(n-2) ) et n ≥ 4 ( 4 + √(n - 2) )
n ≥ 2 et n ≥ 4 donc D = [4 ; +oo[
[tex]4 + \sqrt{n - 2} = n \\ \sqrt{n - 2} = n - 4 \\ n - 2 = (n - 4) {}^{2} \\ n - 2 = n {}^{2} - 8n + 16 \\ n {}^{2} - 9n + 18 = 0 \\ n {}^{2} - 3n - 6n + 18 = 0 \\ n(n - 3) - 6(n - 3) = 0 \\ (n - 3)(n - 6) = 0[/tex]
n - 3 = 0 ou n - 6 = 0
n = 3 ou n = 6
n = 6 car n ≥ 4
S = { 6 }
