AB: Lektion Brüche (Teil 3)

1.

Addiere die folgenden Brüche. Versuche außerdem, das Ergebnis zu kürzen und es als gemischte Zahl zu schreiben:

a)

\( \frac{2}{3} + \frac{4}{3} = \) \( \bbox[#e1ffc1,5px]{ \frac{2+4}{3} = \frac{6}{3} = 2 } \)

b)

\( \frac{1}{2} + \frac{4}{2} = \) \( \bbox[#e1ffc1,5px]{ \frac{1+4}{2} = \frac{5}{2} = 2 \frac{1}{2} } \)

c)

\( \frac{1}{3} + \frac{1}{2} = \) \( \bbox[#e1ffc1,5px]{ \frac{1·2}{3·2} + \frac{1·3}{2·3} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} } \)

d)

\( \frac{11}{4} + \frac{12}{6} = \) \( \bbox[#e1ffc1,5px]{ \frac{11·3}{4·3} + \frac{12·2}{6·2} \\ = \frac{33}{12} + \frac{24}{12} = \frac{57}{12} \\ = \frac{57:3}{12:3} = \frac{19}{4} \\ = 4 \frac{3}{4} } \)

e)

\( \frac{9}{2} + \frac{1}{21} + \frac{4}{14} = \) \( \bbox[#e1ffc1,5px]{ \frac{9·21}{2·21} + \frac{1·2}{21·2} + \frac{4·3}{14·3} \\ = \frac{189}{42} + \frac{2}{42} + \frac{12}{42} \\ = \frac{189 + 2 + 12}{42} = \frac{203}{42} \\ = \frac{168+35}{42} = \frac{168}{42} + \frac{35}{42} \\ = 4 \frac{35}{42} = 4 \frac{35:7}{42:7} \\ = 4 \frac{5}{6} } \)

f)

\( \frac{1}{8} + \frac{1}{4} + \frac{1}{3} = \) \( \bbox[#e1ffc1,5px]{ \frac{1·3}{8·3} + \frac{1·6}{4·6} + \frac{1·8}{3·8} \\ = \frac{3}{24} + \frac{6}{24} + \frac{8}{24} \\ = \frac{3+6+8}{24} = \frac{17}{24} } \)

g)

\( \frac{19}{8} + \frac{3}{5} + 8\frac{1}{2} + \frac{21}{40} = \) \( \bbox[#e1ffc1,5px]{ \frac{19}{8} + \frac{3}{5} + \frac{16+1}{2} + \frac{21}{40} \\ = \frac{19}{8} + \frac{3}{5} + \frac{17}{2} + \frac{21}{40} \\ = \frac{19·5}{8·5} + \frac{3·8}{5·8} + \frac{17·20}{2·20} + \frac{21}{40} \\ = \frac{95}{40} + \frac{24}{40} + \frac{340}{40} + \frac{21}{40} \\ = \frac{95 + 24 + 340 + 21}{40} \\ = \frac{480}{40} = 12 } \)

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