Question

Сравните числа (не пользуясь калькулятором)

Answer 1

$a=\dfrac{1+\sqrt5}{1-\sqrt5}=\dfrac{(1+\sqrt5)^2}{(1-\sqrt5)(1+\sqrt5)}=\dfrac{1+2\sqrt5+5}{1-5}=-\dfrac{3}{2}-\dfrac{\sqrt5}{2}\\ b=\dfrac{2}{1-\sqrt3}=\dfrac{2(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)}=-1-\sqrt3\\ a+1=-\dfrac{1}{2}-\dfrac{\sqrt5}{2}\;\;\;\;\;\;\;\;\;\;\;\;b+1=-\sqrt3\\ (a+1)^2=\dfrac{1}{4}+\dfrac{\sqrt5}{2}+\dfrac{5}{4}=\dfrac{3}{2}+\dfrac{\sqrt5}{2}\\ (b+1)^2=3=\dfrac{3}{2}+\dfrac{3}{2}\\ (a+1)^2- (b+1)^2=\dfrac{\sqrt5}{2}-\dfrac{3}{2}=\dfrac{\sqrt5-\sqrt9}{2}<0=> (a+1)^2< (b+1)^2$

$a+1<0,\:b+1<0=>a+1>b+1=>a>b$

2)

$a^2=2018+2\sqrt{2018\cdot2020}+2020=2019*2+2\sqrt{(2019-1)(2019+1)}=(\sqrt{2019})^2+2\sqrt{2019^2-1^2}+(\sqrt{2019})^2<(\sqrt{2019})^2+2\sqrt{2019^2}+(\sqrt{2019})^2=(\sqrt{2019})^2+2(\sqrt{2019})^2+(\sqrt{2019})^2=(2\sqrt{2019})^2=b^2\\ a,b>0=>a<b$