Odpowiedź:
[tex]a)\ \ 4^{-\frac{1}{2}}+27^{-\frac{1}{3}}=(\frac{1}{4})^{\frac{1}{2}}+(\frac{1}{27})^{\frac{1}{3}}=\sqrt{\frac{1}{4}}+\sqrt[3]{\frac{1}{27}}=\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\\\\\\b)\ \ 5^{-\frac{5}{4}}\cdot5^{\frac{4}{3}}=5^{-\frac{5}{4}+\frac{4}{3}}=5^{-\frac{15}{12}+\frac{16}{12}}=5^{\frac{1}{12}}[/tex]
[tex]c)\ \ 9^{\frac{3}{10}}:27^{-\frac{4}{5}}=(3^2)^{\frac{3}{10}}:(3^3)^{-\frac{4}{5}}=3^{\frac{3}{5}}:3^{-\frac{12}{5}}=3^{\frac{3}{5}-(-\frac{12}{5})}=3^{\frac{3}{5}+\frac{12}{5}}=3^{\frac{15}{5}}=\\\\=3^3=27\\\\\\d)\ \ log_{\frac{1}{3}}27=log_{3^{-1}}3^3=\frac{3}{-1}\cdot log_{3}3=-3\cdot1=-3\\\\\\e)\ \ log_{27}3\sqrt{3}=log_{3^3}3^{\frac{3}{2}}=\dfrac{\frac{3}{2}}{3}\cdot log_{3}3=\frac{\not3}{2}\cdot\frac{1}{\not3}\cdot1=\frac{1}{2}[/tex]
