Odpowiedź:
[tex]a)\ \ \dfrac{2^0+2^{-2}}{(\frac{2}{3})^{-2}-5\cdot(-2)^{-2}+(\frac{1}{2})^{-2}}+\frac{3}{4}=\dfrac{1+(\frac{1}{2})^2}{(\frac{3}{2})^2-5\cdot(-\frac{1}{2})^2+2^2}+\frac{3}{4}=\\\\\\=\dfrac{1+\frac{1}{4}}{\frac{9}{4}-5\cdot\frac{1}{4}+4}+\frac{3}{4}=\dfrac{\frac{5}{4}}{\frac{9}{4}-\frac{5}{4}+4}+\frac{3}{4}=\dfrac{\frac{5}{4}}{1+4}+\frac{3}{4}=\dfrac{\frac{5}{4}}{5}+\frac{3}{4}=\frac{5}{4}:5+\frac{3}{4}=[/tex]
[tex]=\frac{\not5}{4}\cdot\frac{1}{\not5}+\frac{3}{4}=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1[/tex]
[tex]b)\ \ (\frac{1}{2})^{-5}\cdot8^{-2}+(-64^{-1})^{-2}\cdot2^{-13}-0,2^{-3}\cdot125^{-1}=\\\\=2^5\cdot(2^3)^{-2}+(-64)^{2}\cdot2^{-13}-(\frac{1}{5})^{-3}\cdot(5^3)^{-1}=2^5\cdot2^{-6}+64^2\cdot2^{-13}-5^3\cdot5^{-3}=\\\\=2^{5-6}+(2^6)^2\cdot2^{-13}-5^{3-3}=2^{-1}+2^{12}\cdot2^{-13}-5^0=2^{-1}+2^{12-13}-1=\\\\=2^{-1}+2^{-1}-1=\frac{1}{2}+\frac{1}{2}-1=1-1=0[/tex]
[tex]c)\ \ 3\cdot[5-(\frac{1}{2})^{-1}]^2\cdot[(-\frac{2}{3})^{-3}+3\cdot2^{-3}]^{-2}=3\cdot(5-2)^2\cdot[(-\frac{3}{2})^3+3\cdot(\frac{1}{2})^3]^{-2}=\\\\=3\cdot3^2\cdot(-\frac{27}{8}+3\cdot\frac{1}{8})^{-2}=3\cdot9\cdot(-\frac{27}{8}+\frac{3}{8})^{-2}=27\cdot(-\frac{24}{8})^{-2}=27\cdot(-3)^{-2}=\\\\=27\cdot(-\frac{1}{3})^2=\not27^3\cdot\frac{1}{\not9_{1}}=3[/tex]
[tex]d)\ \ \dfrac{[(\frac{1}{4^{-1}}+\frac{3}{9^{-1}})-(\frac{1}{5^{-2}}-\frac{2}{3^{-2}})]\cdot(5^0-5^{-2})^{-1}}{(2\cdot5^{-1}-5^{-2})^{-1}}=\\\\\\=\dfrac{[(1\cdot4^1+3\cdot9^1)-(1\cdot5^2-2\cdot3^2)]\cdot[1-(\frac{1}{5})^2]^{-1}}{[2\cdot\frac{1}{5}-(\frac{1}{5})^2]^{-1}}=\\\\\\=\dfrac{[(1\cdot4+3\cdot9)-(1\cdot25-2\cdot9)]\cdot(1-\frac{1}{25})^{-1}}{(\frac{2}{5}-\frac{1}{25})^{-1}}=\dfrac{[(4+27)-(25-18)]\cdot(\frac{25}{25}-\frac{1}{25})^{-1}}{(\frac{10}{25}-\frac{1}{25})^{-1}}=[/tex]
[tex]=\dfrac{(31-7)\cdot(\frac{24}{25})^{-1}}{(\frac{9}{25})^{-1}}=\dfrac{\not24\cdot\frac{25}{\not24}}{\frac{25}{9}}=\dfrac{25}{\frac{25}{9}}=25:\frac{25}{9}=\not25\cdot\frac{9}{\not25}=9[/tex]
