Odpowiedź:
[tex]a)\ \ (\frac{1}{2})^2-3^2\cdot(\frac{1}{3})^2+4^3\cdot(\frac{1}{4})^2=\frac{1}{4}-\not9^1\cdot\frac{1}{\not9_{1}}+\not64^4\cdot\frac{1}{\not16_{1}}=\frac{1}{4}-1+4=\frac{1}{4}+3=3\frac{1}{4}\\\\\\b)\ \ 3\cdot(\frac{1}{2})^3-2\cdot(\frac{1}{3})^2+5\cdot(\frac{1}{4})^2=3\cdot\frac{1}{8}-2\cdot\frac{1}{9}+5\cdot\frac{1}{16}=\frac{3}{8}-\frac{2}{9}+\frac{5}{16}=\frac{54}{144}-\frac{32}{144}+\frac{45}{144}=\\\\=\frac{22}{144}+\frac{45}{144}=\frac{67}{144}[/tex]
[tex]c)\ \ (\frac{1}{2})^2-(\frac{2}{3})^2+(\frac{3}{4})^2=\frac{1}{4}-\frac{4}{9}+\frac{9}{16}=\frac{36}{144}-\frac{64}{144}+\frac{81}{144}=\frac{117}{144}-\frac{64}{144}=\frac{53}{144}\\\\\\d)\ \ (0,1)^2+(0,2)^2-(0,3)^2=0,01+0,04-0,09=0,05-0,09=-0,04[/tex]
[tex]lub\\\\(0,1)^2+(0,2)^2-(0,3)^2=(\frac{1}{10})^2+(\frac{2}{10})^2-(\frac{3}{10})^2=\frac{1}{100}+\frac{4}{100}-\frac{9}{100}=\frac{5}{100}-\frac{9}{100}=\\\\=-\frac{4}{100}=-\frac{1}{25}[/tex]
