Question

Ajuutooor
Determinati k apartine lui Z pentru care numarul a =k/(2+√2)-(2-√2)²-√(3+√8)² este numar intreg

Answer 1

[tex]a=\frac{k}{2+\sqrt{2}}-(2-\sqrt{2})^2 - \sqrt{(3+\sqrt{8})^2} \\ a=\frac{k(2-\sqrt{2})}{2}-(4-4\sqrt{2}+2)-|3+2\sqrt{2}| \\ a=\frac{k(2-\sqrt{2})}{2}-6+4\sqrt{2}-3-2\sqrt{2} \\ a=\frac{k(2-\sqrt{2})}{2}-9+2\sqrt{2} [/tex]
Pentru k = 4, a devine:
$a=\frac{4(2-\sqrt{2})}{2} - 9 +2\sqrt{2} \\ a = 2(2-\sqrt{2})-9+2\sqrt{2} \\ a = 4 - 2\sqrt{2}-9+2\sqrt{2} =\ \textgreater \ a = -5 \in Z$