Question

Сравните числа (не пользуясь калькулятором)

Answer 1

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

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

2)

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

Answer 2

Поможешь ещё, пожалуйста?