• LK Nr. 1 - Brüche und Bruchrechnen
  • Christian Leeser
  • 30.06.2020
  • Mittlere Reife
  • Mathematik
  • 6
Um die Lizenzinformationen zu sehen, klicken Sie bitte den gewünschten Inhalt an.
1
Gib je­weils den Bruch­teil der dunk­le­ren Teile an.
4 / 4
Lösung
a) 412\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{4}{12} b) 39\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{3}{9}

c) 26\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{2}{6} d) 13\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{1}{3}
2
Er­stel­le zu den auf­ge­führ­ten Brü­chen je­weils eine Figur.
4 / 4
  • a) 12\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{1}{2}
  • b) 34\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{3}{4}
  • c) 57\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{5}{7}
  • d) 610\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{6}{10}
3
Wand­le in einen un­ech­ten Bruch um.
8 / 8
  • a) 312\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 3\frac{1}{2}

  • b) 734\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 7 \frac{3}{4}

  • c) 657\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 6\frac{5}{7}

  • d) 7610\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 7\frac{6}{10}
Lösung
a) 72\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{7}{2} b) 314\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{31}{4}
c) 477\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{47}{7} d) 7610\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{76}{10}
4
Wand­le in einen ge­misch­te Schreib­wei­se um.
8 / 8
  • a) 112\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{11}{2}

  • b) 134\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{13}{4}

  • c) 657\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{65}{7}

  • d) 9910\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{99}{10}
Lösung
a) 512\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 5 \frac{1}{2} b) 314\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 3 \frac{1}{4}
c) 927\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 9 \frac{2}{7} d) 9910\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} 9 \frac{9}{10}
Note
/ 24
Unterschrift