[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.
