a) 27^{4\sqrt7}* 81^{0,5} -3^{\sqrt7}=(3^3)^3\sqrt7=(3^3)^{4\sqrt7}*(3^4)^{0,5}-3^{\sqrt7}=
=3^{12\sqrt7}*3^2-3^{\sqrt7}=(3^2)^{6\sqrt7}-9-3\sqrt7=9^{6\sqrt7}-3\sqrt7
b)
16^{5\sqrt6} : 4^2^{\sqrt[3]{750}}=(2^4)^{5\sqrt6}:(2^2)^{2\sqrt[3]{125*6}}=2^{20\sqrt6}:2^{4*5\sqrt[3]6}=
=2^{20\sqrt6-20\sqrt[3]6}=2^{20(\sqrt6-\sqrt[3]6)}
I pierwiastek jest kwadratowy, czy sześcienny?
c)
[(3 - \sqrt5)^{\sqrt2}(3 + \sqrt5)^{\sqrt2}]^{\sqrt2}=[(3 - \sqrt5)(3 + \sqrt5)^{\sqrt2}]^{\sqrt2}=[(9-5)^\{\sqrt2}]^{\sqrt2}=
=4^{\sqrt2*\sqrt2}=4^2=16