Question

Aratati ca numarul
a= I \sqrt{5} -3I/ + \frac{4}{ 3-\sqrt{5} }  este intreg

Answer 1

a=|√5-3| + 4/(3-√5)=3-√5+4/(3-√5)=[(3-√5)^2+4]/(3-√5)
a=(18-6√5)/(3-√5)=6(3-√5)/(3-√5)=6

Answer 2

$\displaystyle a= | \sqrt{5} -3|+ \frac{4}{ 3-\sqrt{5}} \\ a=3- \sqrt{5} + \frac{4\left(3+ \sqrt{5} \right)}{9-5} \\ a= 3- \sqrt{5} + \frac{4\left(3+ \sqrt{5}\right) }{4} \\ a=3- \sqrt{5} +3+ \sqrt{5} \\ a=6 \in Z$