Szczegółowe wyjaśnienie:
Doprowadzamy potęgi do tych samych podstaw.
[tex](\frac{1}{3} )^{-4} *3^{-3} :3^{2} =(\frac{3}{1} )^{4}*3^{-3} :3^{2} =3^{4} *3^{-3} :3^{2} =3^{4+(-3)} :3^{2}=3^{1} :3^{2} =[/tex]
[tex]=3^{1-2} =3^{-1} =\frac{1}{3}[/tex]
==================================================
[tex]2^{-5} *(\frac{1}{\sqrt{2} } )^{-4} *(\sqrt{8} )^{6} =2^{-5} *(\frac{\sqrt{2} }{1} )^{4} *((\sqrt{8})^{2})^{3} =[/tex]
[tex]=2^{-5} *((\sqrt{2} )^{2} )^{2} *(8)^{3} =[/tex]
[tex]= 2^{-5} *2^{2} *(2^{3} )^{3} =2^{-5} *2^{2} *2^{9} =2^{-5+2+9}=2^{6} =64[/tex]
Korzystamy z własności potęg:
[tex]a^{-1} =(\frac{1}{a} )^{1}[/tex]
[tex]a^{m} *a^{n} =a^{m+n}[/tex]
[tex]a^{m} :a^{n} =a^{m-n}[/tex]
[tex](a^{m} )^{n} =a^{m*n}[/tex]
[tex](\sqrt{a})^{2} =a[/tex]