Question

Demonstreza ca a este natural daca a=√(15-3√11)la patrat+ √(3-√99)la patrat va rog dau coroana exercitiul pe deplin va rog urgent

Answer 1

$a=\sqrt{(15-3\sqrt{11})^2}+\sqrt{(3-\sqrt{99})^2}=|15-3\sqrt{11}|+|3-\sqrt{99}|$

$225>99\Rightarrow\sqrt{225}>\sqrt{99},\ sau\ 15>\sqrt{9\cdot 11},\ sau\ 15>3\sqrt{11},\\deci\ |15-3\sqrt{11}|=15-3\sqrt{11};\\\\99>9\Rightarrow\sqrt{99}>\sqrt{9},\ sau\ \sqrt{9\cdot 11}>3,\ sau\ 3\sqrt{11}>3,\\deci\ |3-\sqrt{99}|=-(3-\sqrt{99})=\sqrt{99}-3=3\sqrt{11}-3.$
Numărul a devine deci:
$a=15-3\sqrt{11}+3\sqrt{11}-3=12\in\mathbb{N}.$

Nu a fost chiar așa de greu, nu ?
Green eyes.