Zadanie 1
[tex]\dfrac{4^3\cdot16^{\frac{1}{4}}:\sqrt[5]{32}}{64^{-\frac{3}{4}}\cdot8^{\frac{5}{3}}}=\dfrac{(2^2)^3\cdot(2^4)^{\frac{1}{4}}:(2^5)^{\frac{1}{5}}}{(2^6)^{-\frac{3}{4}}\cdot(2^3)^{\frac{5}{3}}}=\dfrac{2^6\cdot2^1:2^1}{2^{-\frac{9}{2}}\cdot2^5}=\\\\\\=\dfrac{2^{6+1-1}}{2^{-\frac{9}{2}+\frac{10}{2}}}=\dfrac{2^6}{2^{\frac{1}{2}}}=2^{6-\frac{1}{2}}=2^{\frac{12}{2}-\frac{1}{2}}=\boxed{2^{\frac{11}{2}}}[/tex]
Zadanie 2
[tex]\dfrac{12\sqrt{32^2}-\Big(-\dfrac{1}{2}\Big)^{-8}}{2^{\frac{1}{3}}\cdot4^{\frac{1}{3}}+(6^{\frac{1}{2}}\cdot3^{-\frac{1}{2}})^2}=\dfrac{12\cdot\sqrt{(2^5)^2}-(-2)^8}{2^{\frac{1}{3}}\cdot(2^2)^{\frac{1}{3}}+\Big(6^{\frac{1}{2}}\cdot\Big(\dfrac{1}{3}\Big)^\frac{1}{2}\Big)^2}=\\\\\\=\dfrac{12\cdot(2^{10})^{\frac{1}{2}}-2^8}{2^{\frac{1}{3}}\cdot2^{\frac{2}{3}}+\Big(\Big(6\cdot\dfrac{1}{3}\Big)^{\frac{1}{2}}\Big)^2}=\dfrac{12\cdot2^5-2^8}{2^{\frac{1}{3}+\frac{2}{3}}+(2^{\frac{1}{2}})^2}=[/tex]
[tex]=\dfrac{3\cdot2^2\cdot2^5-2^8}{2^1+2^1}=\dfrac{3\cdot2^7-2\cdot2^7}{4}=\dfrac{2^7\cdot(3-2)}{2^2}=\dfrac{2^7}{2^2}=2^{7-2}=\boxed{2^5}[/tex]
Wykorzystaliśmy wzory:
[tex]a^{m+n}=a^m\cdot a^n\\\\(a^m)^n=a^{m\cdot n}\\\\a^{-n}=\Big(\dfrac{1}{a}\Big)^n\\\\a^{\frac{n}{m}}=\sqrt[m]{a^n}\\\\a^n\cdot b^n=(a\cdot b)^n[/tex]
