Oblicz.- Zadanie 6.3: Prosto do matury 1. Zakres podstawowy

Rozwiązanie

$a) (\frac{2}{3} + \frac{1}{4}) : 1 \frac{5}{6} + \frac{1}{2} = (\frac{8}{12} + \frac{3}{12}) : \frac{11}{6} + \frac{1}{2} = \frac{11}{12} : \frac{11}{6} + \frac{1}{2} = \frac{11 ^{1}}{12} \cdot \frac{6}{11 ^{1}} + \frac{1}{2} = \frac{6}{12} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$

$b) [(5 \frac{1}{2} - 3 \frac{1}{4}) : 2 \frac{1}{4} + 7 \frac{5}{8}] : 2 \frac{7}{8} = [(5 \frac{2}{4} - 3 \frac{1}{4}) : \frac{9}{4} + 7 \frac{5}{8}] : \frac{23}{8} = (2 \frac{1}{4} \cdot \frac{4}{9} + 7 \frac{5}{8}) \cdot \frac{8}{23} = (\frac{9}{4} \cdot \frac{4}{9} + 7 \frac{5}{8}) \cdot \frac{8}{23} = (1 + 7 \frac{5}{8}) \cdot \frac{8}{23} = 8 \frac{5}{8} \cdot \frac{8}{23} = \frac{69}{8} \cdot \frac{8}{23} = 3$

$c) \frac{3}{4} - \frac{3}{2} + (2 \frac{1}{2} - 1 \frac{7}{8}) : \frac{1}{8} - 2 \frac{1}{4} = \frac{3}{4} - \frac{6}{4} + (2 \frac{4}{8} - 1 \frac{7}{8}) \cdot \frac{8}{1} - 2 \frac{1}{4} = - \frac{3}{4} + (1 \frac{12}{8} - 1 \frac{7}{8}) \cdot \frac{8}{1} - 2 \frac{1}{4} = - \frac{3}{4} + \frac{5}{8 ^{1}} \cdot \frac{8 ^{1}}{1} - 2 \frac{1}{4} = - \frac{3}{4} + 5 - 2 \frac{1}{4} = 4 \frac{1}{4} - 2 \frac{1}{4} = 2$

$d) (1 \frac{1}{2} + 2 \frac{2}{3}) : \frac{5}{6} - (3 \frac{1}{3} - 1 \frac{1}{4}) : 2 \frac{1}{12} = (1 \frac{3}{6} + 2 \frac{4}{6}) \cdot \frac{6}{5} - (3 \frac{4}{12} - 1 \frac{3}{12}) : 2 \frac{1}{12} = 3 \frac{7}{6} \cdot \frac{6}{5} - 2 \frac{1}{12} : 2 \frac{1}{12} = \frac{25 ^{5}}{6 ^{1}} \cdot \frac{6 ^{1}}{5 ^{1}} - 1 = 5 - 1 = 4$

Dagmara Kowalczuk

Nauczycielka matematyki

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