Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]a) A = \frac{2}{3} + \frac{9}{2} : \frac{9}{4} = \frac{2}{3} + \frac{9}{2} \cdot \frac{4}{9} = \frac{2}{3} + \frac{2}{1} = 2\frac{2}{3} \\B = (\frac{2}{3} + 4\frac{1}{2}) : 2\frac{1}{4} = (\frac{6}{9} + \frac{27}{9}) \cdot \frac{9}{4} = \frac{33}{9} \cdot \frac{9}{4} = \frac{33}{4} = 8\frac{1}{4}\\B-A = 8\frac{1}{4} - 2\frac{2}{3} = 5\frac{7}{12}[/tex]
b)
[tex]A = 1\frac{3}{4}\cdot\frac{5}{7}+\frac{5}{8} = \frac{7}{4}\cdot\frac{5}{7}+\frac{5}{8} =\frac{10}{8} + \frac{5}{8} = \frac{15}{8} = 1\frac{7}{8}\\B = 1\frac{3}{4}\cdot(\frac{5}{7}+\frac{5}{8})=\frac{7}{4}\cdot\frac{5\cdot8+5\cdot7}{56} = \frac{7}{4}\cdot\frac{75}{56}=\frac{75}{32}\\|A-B|=|\frac{15}{8} - \frac{75}{32}| = |\frac{60}{32}-\frac{75}{32}|=|-\frac{15}{32}|=\frac{15}{32}[/tex]
c)
[tex]A=6-\frac{13}{3}\cdot\frac{12}{13} = 6-\frac{12}{3}=6-4=2\\B=(6-4\frac{1}{3})\cdot\frac{12}{13} = 1\frac{2}{3}\cdot\frac{12}{13} = \frac{5\cdot12}{3\cdot13}=\frac{20}{13}\\|A-B|=|2-\frac{20}{13}|=|\frac{26}{13}-\frac{20}{13}|=\frac{6}{13}[/tex]
d)
[tex]A=3\frac{1}{3}\cdot\frac{3}{5}-1\frac{1}{6}=\frac{10}{3}\cdot\frac{3}{5}-\frac{7}{6}=2-\frac{7}{6}=\frac{12}{6}-\frac{7}{6}=\frac{5}{6}\\B=\frac{10}{3}\cdot(\frac{3}{5}-1\frac{1}{6})=\frac{10}{3}\cdot(\frac{18}{30}-\frac{35}{30}) = \frac{10}{3}\cdot(-\frac{17}{30})=-\frac{17}{9}\\|A-B|=|\frac{5}{6}-(-\frac{17}{9})|=|\frac{15}{18}+\frac{34}{18}|=\frac{49}{18}[/tex]
