Question

Aratati ca numarul a=modul de I√5-3I + 4 supra 3-√5 este intreg .

Answer 1

Stim ca |√5-3|=3-√5, deorece 3>√5

Atunci relatia devine:  3-√5+4/(3-√5)= $\frac{(3-\sqrt{5}) ^{2}+4 }{3-\sqrt{5} } =\frac{9-6\sqrt{5}+5+4 }{3-\sqrt{5} }=\frac{18-6\sqrt{5} }{3-\sqrt{5} }  =\frac{6(3-\sqrt{5} )}{3-\sqrt{5} } =6$ ∈Z