[tex]\\ $Voi calcula prin doua cai diferite:\\ \\ \boxed{1} \quad \sqrt{2^8} = (2^8)^{\dfrac{1}{2}} = 2^{\big8\cdot \dfrac{1}{2}} = 2^{\dfrac{8}{2}} = 2^{4} = 16 \\ \\ \boxed{2} \quad \sqrt{2^8} = \sqrt2^{8} = \sqrt2^{2\cdot 4} = \Big(\sqrt{2}^2\Big)^4 = 2^4=16 \\ \boxed{\sqrt{a^b} = \sqrt{a}^b}$ $ \rightarrow $ proprietate.[/tex]
