AB: Kürzen von Brüchen II (Erweitert)

Beim Kürzen werden Zähler und Nenner des Bruches durch die gleiche Zahl dividiert.

Beispiel: \( \frac{30}{50} = \frac{30\color{blue}{:10}}{50\color{blue}{:10}} = \frac{3}{5} \)

Wenn die Kürzzahl nicht bekannt ist, können wir diese berechnen, indem wir den Zähler des ursprünglichen Bruches durch den Zähler des gekürzten Bruches dividieren:

Beispiel: \( \frac{\color{red}{30}}{50} = \frac{30\color{blue}{:x}}{50\color{blue}{:x}} = \frac{\color{red}{3}}{5} \rightarrow \color{blue}{x} = \color{red}{30} : \color{red}{3} = \color{blue}{10} \)

1.

Bestimme die Zahl, die zum gekürzten Bruch führt:

a)

\( \large { \frac{8}{40} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{2}{10} } \) \( \frac{8}{40}^{\bbox[#e1ffc1,5px]{:4}} = \frac{8\color{blue}{:4}}{40\color{blue}{:4}} = \frac{2}{10} \)

b)

\( \large { \frac{3}{9} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{1}{3} } \) \( \frac{3}{9}^{\bbox[#e1ffc1,5px]{:3}} = \frac{3\color{blue}{:3}}{9\color{blue}{:3}} = \frac{1}{3} \)

c)

\( \large { \frac{6}{16} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{3}{8} } \) \( \frac{6}{16}^{\bbox[#e1ffc1,5px]{:2}} = \frac{6\color{blue}{:2}}{16\color{blue}{:2}} = \frac{3}{8} \)

d)

\( \large { \frac{8}{48} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{1}{6} } \) \( \frac{8}{48}^{\bbox[#e1ffc1,5px]{:8}} = \frac{8\color{blue}{:8}}{48\color{blue}{:8}} = \frac{1}{6} \)

e)

\( \large { \frac{12}{84} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{1}{\bbox[5pt, border:0.5pt solid #00C]{ \space }} } \) \( \frac{12}{84}^{\bbox[#e1ffc1,5px]{:12}} = \frac{12\color{blue}{:12}}{84\color{blue}{:12}} = \frac{1}{\bbox[#e1ffc1,5px]{7}} \)

f)

\( \large { \frac{51}{12} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{17}{\bbox[5pt, border:0.5pt solid #00C]{ \space }} } \) \( \frac{51}{12}^{\bbox[#e1ffc1,5px]{:3}} = \frac{51\color{blue}{:3}}{12\color{blue}{:3}} = \frac{17}{\bbox[#e1ffc1,5px]{4}} \)

g)

\( \large { \frac{909}{54} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{101}{\bbox[5pt, border:0.5pt solid #00C]{ \space }} } \) \( \frac{909}{54}^{\bbox[#e1ffc1,5px]{:9}} = \frac{909\color{blue}{:9}}{54\color{blue}{:9}} = \frac{101}{\bbox[#e1ffc1,5px]{6}} \)

h)

\( \large { \frac{155}{25} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{31}{\bbox[5pt, border:0.5pt solid #00C]{ \space }} } \) \( \frac{155}{25}^{\bbox[#e1ffc1,5px]{:5}} = \frac{155\color{blue}{:5}}{25\color{blue}{:5}} = \frac{31}{\bbox[#e1ffc1,5px]{5}} \)

i)

\( \large { \frac{27}{81} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{\bbox[5pt, border:0.5pt solid #00C]{ \space }}{9} } \) \( \frac{27}{81}^{\bbox[#e1ffc1,5px]{:9}} = \frac{27\color{blue}{:9}}{81\color{blue}{:9}} = \frac{\bbox[#e1ffc1,5px]{3}}{9} \)

j)

\( \large { \frac{30}{195} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{\bbox[5pt, border:0.5pt solid #00C]{ \space }}{13} } \) \( \frac{30}{195}^{\bbox[#e1ffc1,5px]{:15}} = \frac{30\color{blue}{:15}}{195\color{blue}{:15}} = \frac{\bbox[#e1ffc1,5px]{2}}{13} \)

k)

\( \large { \frac{175}{205} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{\bbox[5pt, border:0.5pt solid #00C]{ \space }}{41} } \) \( \frac{175}{205}^{\bbox[#e1ffc1,5px]{:5}} = \frac{175\color{blue}{:5}}{205\color{blue}{:5}} = \frac{\bbox[#e1ffc1,5px]{35}}{41} \)

l)

\( \large { \frac{210}{1020} ^ {\color{#00F}{:\,} \bbox[3pt, border:0.5pt solid #00F]{ \color{#FFF}{x} } } = \frac{\bbox[5pt, border:0.5pt solid #00C]{ \space }}{102} } \) \( \frac{210}{1020}^{\bbox[#e1ffc1,5px]{:10}} = \frac{210\color{blue}{:10}}{1020\color{blue}{:10}} = \frac{\bbox[#e1ffc1,5px]{21}}{102} \)

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