[tex]a=\sqrt{(15-3\sqrt{11})^2}+\sqrt{(3-\sqrt{99})^2}=|15-3\sqrt{11}|+|3-\sqrt{99}|[/tex]
[tex]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.[/tex]
Numărul a devine deci:
[tex]a=15-3\sqrt{11}+3\sqrt{11}-3=12\in\mathbb{N}.[/tex]
Nu a fost chiar așa de greu, nu ?
Green eyes.
