Question

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

Answer 1

Intrucat √5-3 <0 ,cand vom explicita modulul vom avea 3-√5(acesta fiind un numar pozitiv).Astfel ecuatia noastra va lua forma:$3- \sqrt{5}$)+($\frac{4}{3- \sqrt{5}}$.Aducem la acelasi numitor,amplificand cu $3- \sqrt{5}$ si vom obtine:
$\frac{(3- \sqrt{5}) ^{2} +4 }{3- \sqrt{5} }= \frac{9+5-6 \sqrt{5} +4}{3- \sqrt{5} } = \frac{18-6 \sqrt{5} }{3-\sqrt{5}}= \frac{6(3-\sqrt{5})}{3-\sqrt{5}} =6$ ∈Z

Answer 2

deoarece $\sqrt{5}-3<0\Rightarrow|\sqrt5-3|=-(\sqrt5-3)=3-\sqrt5$
Inlocuim si rationalizam fractia
$3-\sqrt5+\dfrac{4(3+\sqrt5)}{3^2-(\sqrt5)^2}=3-\sqrt5+\dfrac{4(3+\sqrt5)}{4}=$
$=3-\sqrt5+3+\sqrt5=6\in Z$