Locala Bihor (12.02.2011), P1

[mihai miculita]

Calculati pe $x\in\mathbb{R}$, din: a). $\frac{\sqrt{102,01}+x}{2x}$$=\frac{7,1}{\sqrt{2^8+\sqrt{100^2-80^2}}}.$ b). $\frac{\sqrt{2}+\sqrt{15}+\sqrt{5}+\sqrt{7}+\sqrt{6}+\sgrt{21}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}-x=1.$