a)
\frac{8^{\sqrt2}}{2^{\sqrt2}}=(\frac{8}{2})^{\sqrt2}=4^{\sqrt2}=2^{2\sqrt2}
b)
2^{1-\sqrt2} \cdot 2^{1+\sqrt2}=2^{(1-\sqrt2)(1+\sqrt2)}=2^{1-(\sqrt2)^2}=2^{1-2}=2^{-1}=\frac{1}{2}
c)
[(\frac{1}{3})^{1-\sqrt3}]^{1-\sqrt3}=(\frac{1}{3})^{(1-\sqrt3)^2}=(\frac{1}{3})^{1-2\sqrt3+3}=(\frac{1}{3})^{4-2\sqrt3}
II
{[(\frac{1}{3})^{1-\sqrt3}]^{1+\sqrt3}=(\frac{1}{3})^{(1-\sqrt3)(1+\sqrt3)}=(\frac{1}{3})^{1-(\sqrt3)^2}=(\frac{1}{3})^{1-3}=(\frac{1}{3})^{1-3}=(\frac{1}{3})^{-2}=3^2=9}
e)
(3^{\sqrt7-1})^{\sqrt7+1}=3^{(\sqrt7-1)(\sqrt7+1)}=3^{(\sqrt7)^2-1}=3^{7-1}=3^6
f)
(2^{\sqrt2+\sqrt5})^{\sqrt2+\sqrt5}=2^{(\sqrt2+\sqrt5)^2}=2^{2+\sqrt{10}+5}=2^{7+\sqrt{10}}
lub
{(2^{\sqrt2-\sqrt5})^{\sqrt2+\sqrt5}=2^{(\sqrt2-\sqrt5)(\sqrt2+\sqrt5)}=2^{(\sqrt2)^2-(\sqrt5)^2}=2^{2-5}=2^{-3}=(\frac{1}{2})^3=\frac{1}{8}}
Zastosowany wzór skróconego mnożenia
a^2-b^2=(a-b)(a+b)