Odpowiedź:
[tex](\frac{1}{2})^3+\frac{3}{7}*2,1-\frac{2}{3}*(2,25:2,5)=\frac{1}{8}+\frac{3}{\not7_{1} }*\frac{\not21^3}{10}-\frac{2}{3}*0,9=\frac{1}{8}+\frac{9}{10}-\frac{\not2^1}{\not3_{1} }*\frac{\not9^3}{\not10_{5}}=\\\\=\frac{1}{8}+\frac{9}{10}-\frac{3}{5}=\frac{5}{40}+\frac{36}{40}-\frac{24}{40}=\frac{17}{40}[/tex]
[tex]je\'sli\ \ bedzie\ \ (\frac{1}{3})^3[/tex]
[tex](\frac{1}{3})^3+\frac{3}{7}*2,1-\frac{2}{3}*(2,25:2,5)=\frac{1}{27}+\frac{3}{\not7_{1} }*\frac{\not21^3}{10}-\frac{2}{3}*0,9=\frac{1}{27}+\frac{9}{10}-\frac{\not2^1}{\not3_{1} }*\frac{\not9^3}{\not10_{5}}=\\\\=\frac{1}{27}+\frac{9}{10}-\frac{3}{5}=\frac{10}{270}+\frac{243}{270}-\frac{162}{270}=\frac{91}{270}[/tex]
