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Terme mit rationalen Zahlen

Aufgabe 1

a)
\(\begin{array}[t]{rll} \dfrac{3}{5}+\dfrac{2}{15}-\dfrac{4}{10} &=& &\\[5pt] \end{array}\)
b)
\(\begin{array}[t]{rll} \left(-\dfrac{4}{6}\right) - \left(-\dfrac{3}{2}\right) + \dfrac {2}{3} &=& &\\[5pt] \end{array}\)
c)
\(\begin{array}[t]{rll} \dfrac{7}{9} - \dfrac{8}{9} - \dfrac {2}{3} &=& &\\[5pt] \end{array}\)
d)
\(\begin{array}[t]{rll} \left(-\dfrac{6}{7}\right) - \dfrac{3}{14}+ \left(-\dfrac{4}{28}\right) &=& &\\[5pt] \end{array}\)
e)
\(\begin{array}[t]{rll} \dfrac{5}{8}+ \dfrac {16}{24}- \left(-\dfrac{3}{8}\right) + \dfrac {18}{24} &=& &\\[5pt] \end{array}\)
f)
\(\begin{array}[t]{rll} \left(-\dfrac{6}{5}\right) - \dfrac {16} {15} + \left(-\dfrac {22}{10}\right) - \left(-\dfrac{15}{25}\right) &=& &\\[5pt] \end{array}\)

Aufgabe 2

a)
\(\begin{array}[t]{rll} \dfrac{3}{5}\cdot \dfrac {7}{6}: \dfrac {5}{2} &=& &\\[5pt] \end{array}\)
b)
\(\begin{array}[t]{rll} \left(-\dfrac{8}{9}\right) : \dfrac{2}{5} : \dfrac {3}{4} &=& &\\[5pt] \end{array}\)
c)
\(\begin{array}[t]{rll} \left(-\dfrac {7}{4}\right) : \left(-\dfrac{3}{8}\right) \cdot \dfrac {2}{9} &=& &\\[5pt] \end{array}\)
d)
\(\begin{array}[t]{rll} \dfrac{10}{3} \cdot \dfrac {4}{6} : \left(-\dfrac {3}{2}\right) &=& &\\[5pt] \end{array}\)
e)
\(\begin{array}[t]{rll} \dfrac{3}{9} : \left(-\dfrac{2}{3}\right) \cdot \left(-\dfrac{5}{2}\right) &=& &\\[5pt] \end{array}\)
f)
\(\begin{array}[t]{rll} \left(-\dfrac{4}{5}\right) \cdot \left(-\dfrac{2}{3}\right) : \left(-\dfrac{6}{9}\right) &=& &\\[5pt] \end{array}\)

Aufgabe 3

Denke an "Punkt vor Strichrechnung".
a)
\(\begin{array}[t]{rll} \left(\dfrac{6}{7} - \dfrac {3}{7}\right) : \left(-\dfrac{5}{8}\right) &=& &\\[5pt] \end{array}\)
b)
\(\begin{array}[t]{rll} \dfrac{7}{9}\cdot \dfrac{2}{3}+ \left(-\dfrac {5}{27}\right) &=& &\\[5pt] \end{array}\)
c)
\(\begin{array}[t]{rll} \dfrac{3}{16}- \left(-\dfrac {4}{16}\right)\cdot \dfrac {3}{2} &=& &\\[5pt] \end{array}\)
d)
\(\begin{array}[t]{rll} \dfrac{6}{9} + \dfrac {5}{12} -\dfrac {3}{4} \cdot \dfrac {6}{3} &=& &\\[5pt] \end{array}\)
e)
\(\begin{array}[t]{rll} \left(-\dfrac{7}{3}\cdot\dfrac{3}{4}\right):2 -\dfrac{5}{6} &=& &\\[5pt] \end{array}\)
f)
\(\begin{array}[t]{rll} \left(-\dfrac{8}{9}\right)\cdot \left(-\dfrac{1}{2}\right)+ \dfrac {5}{6}:\left(-\dfrac{3}{4}\right) &=& &\\[5pt] \end{array}\)

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