Multumesc anticipat!
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Răspuns:
asa este!!
Explicație pas cu pas:
2007*2009*2015*2017+64=(2012-3) (2012+3) (2012-5) (2012+5)+64=
(2012²-9) (2012²-25)+64=
2012^4-34*2012²+225+64=
2012^4-34*2012²+289=
(2012²-17)²care e p.p
deci √(2012²-17)²=2012²-17∈N
[tex]\ir Not\breve{a}m\ 2007=a,\ iar\ produsul\ de\ sub\ radical\ devine:\\ \\ a(a+2)(a+8)(a+10)=[a(a+10)][(a+2)(a+8)]=(a^2+10a)(a^2+10a+16)\\ \\ Not\breve{a}m\ a^2+10a = b,\ iar\ expresia\ de\ sub\ radical\ devine:\\ \\ b(b+16)+64=b^2+16b+8^2 =(b+8)^2[/tex]
Prin urmare, sub radical avem un pătrat perfect, deci A∈ ℕ.