Wykonaj działania.- Zadanie 1: Matematyka poznać, zrozumieć 3. Zakres podstawowy. Po gimnazjum

Rozwiązanie

$a) \frac{( 12 \frac{3}{4} - 8 \frac{3}{8} ) \cdot 0 , 3}{0 , 2} = \frac{( 12 \frac{6}{8} - 8 \frac{3}{8} ) \cdot \frac{3}{10}}{\frac{2}{10}} = 4 \frac{3}{8} \cdot \frac{3}{10} \cdot \frac{10}{2} = \frac{35}{8} \cdot \frac{3}{2} = \frac{105}{16} = 6 \frac{9}{16}$

$b) \frac{( 9 \frac{5}{24} - 6 \frac{7}{20} ) : 2 \frac{1}{3} + 0 , 6}{0 , 2} = \frac{( 9 \frac{25}{120} - 6 \frac{42}{120} ) : \frac{7}{3} + \frac{6}{10}}{\frac{2}{10}} = ((8 \frac{145}{120} - 6 \frac{42}{120}) \cdot \frac{3}{7} + \frac{3}{5}) \cdot \frac{10}{2} = (2 \frac{103}{120} \cdot \frac{3}{7} + \frac{3}{5}) \cdot 5 = (\frac{343}{120} \cdot \frac{3}{7} + \frac{3}{5}) \cdot 5 = (\frac{49}{40} + \frac{3}{5}) \cdot 5 = \frac{49}{8} + 3 = 6 \frac{1}{8} + 3 = 9 \frac{1}{5}$

$c) \frac{2 \frac{5}{8} - \frac{2}{3} \cdot 2 \frac{5}{14}}{( 3 \frac{1}{12} + 4 , 375 ) : 19 \frac{8}{9}} = \frac{2 \frac{5}{8} - \frac{2}{3} \cdot \frac{33}{14}}{( 3 \frac{1}{12} + 4 \frac{3}{8} ) : \frac{179}{9}} = \frac{2 \frac{5}{8} - \frac{11}{7}}{( 3 \frac{2}{24} + 4 \frac{9}{24} ) \cdot \frac{9}{179}} = \frac{2 \frac{5}{8} - 1 \frac{4}{7}}{7 \frac{11}{24} \cdot \frac{9}{179}} = \frac{2 \frac{35}{56} - 1 \frac{32}{56}}{\frac{179}{24} \cdot \frac{9}{179}} = \frac{1 \frac{3}{56}}{\frac{3}{8}} = \frac{59}{56} \cdot \frac{8}{3} = \frac{59}{21} = 2 \frac{17}{21}$

$d) \frac{1 + \frac{1}{2} \cdot \frac{1}{0 , 25}}{6 - \frac{46}{1 + 2 , 2 \cdot 10}} = \frac{1 + \frac{1}{2} \cdot \frac{1}{\frac{1}{4}}}{6 - \frac{46}{1 + 22}} = \frac{1 + \frac{1}{2} \cdot 4}{6 - \frac{46}{23}} = \frac{1 + 2}{6 - 2} = \frac{3}{4}$

$e) (1, 5291 - \frac{14 , 53662}{3 - 0 , 095} \cdot 0, 305) : 0, 12 = (1, 5291 - \frac{14 , 53662}{2 , 905} \cdot 0, 305) : 0, 12 = (1, 5291 - 5, 004 \cdot 0, 305) : 0, 12 = (1, 5291 - 1, 52622) : 0, 12 = 0, 00288 : 0, 12 = 0, 024 = \frac{24}{1000} = \frac{3}{125}$

$f) \frac{5 \frac{10}{63} - 4 \frac{25}{84}}{( 2 \frac{1}{3} - 1 \frac{1}{9} ) : 4 - \frac{3}{4}} : 31 = \frac{5 \frac{40}{252} - 4 \frac{75}{252}}{( 2 \frac{3}{9} - 1 \frac{1}{9} ) \cdot \frac{1}{4} - \frac{3}{4}} \cdot \frac{1}{31} = \frac{4 \frac{292}{252} - 4 \frac{75}{252}}{1 \frac{2}{9} \cdot \frac{1}{4} - \frac{3}{4}} \cdot \frac{1}{31} = \frac{\frac{217}{252}}{\frac{11}{9} \cdot \frac{1}{4} - \frac{3}{4}} \cdot \frac{1}{31} = \frac{\frac{217}{252} \cdot \frac{1}{31}}{\frac{11}{36} - \frac{27}{36}} = \frac{\frac{7}{252}}{- \frac{16}{36}} = \frac{\frac{1}{36}}{- \frac{4}{9}} = \frac{1}{36} \cdot (- \frac{9}{4}) = - \frac{1}{16}$

Dagmara Kowalczuk

Nauczycielka matematyki

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