Kapitel 1: Bruchrechnung (S. 11)

Aufgabe 1

(a) \(\displaystyle\left(\frac{4}{5}\right)^2=\frac{4^2}{5^2}=\frac{16}{25}\)

(b) \(\displaystyle\left(\frac{8}{3}\right)^{-1}=\left(\frac{3}{8}\right)^1=\frac{3}{8}\)

(c) \(\displaystyle\left(-\frac{5}{2}\right)^3=\left(\frac{-5}{2}\right)^3=\frac{(-5)^3}{2^3}\) \(\displaystyle=\frac{-125}{8}=-\frac{125}{8}\)

Aufgabe 2

(a) \(\displaystyle\left(\frac{2}{5}\right)^{-1}-\frac{7}{4}=\left(\frac{5}{2}\right)^{1}-\frac{7}{4}=\frac{5}{2}-\frac{7}{4}\) \(\displaystyle=\frac{10}{4}-\frac{7}{4}=\frac{3}{4}=0,75\)

(b) \(\displaystyle\frac{2}{9}+1,7=\frac{2}{9}+\frac{17}{10}=\frac{20}{90}+\frac{153}{90}\) \(\displaystyle=\frac{173}{90}=1,9222…=1,9\overline{2}\)

(c) \(\displaystyle\left(\frac{6}{8}\right)^3-0,625^2=\left(\frac{3}{4}\right)^3-\left(\frac{625}{1000}\right)^2=\frac{3^3}{4^3}-\left(\frac{5}{8}\right)^2\) \(\displaystyle=\frac{27}{64}-\frac{5^2}{8^2}=\frac{27}{64}-\frac{25}{64}=\frac{2}{64}\) \(\displaystyle=\frac{1}{32}=0,03125\)

Aufgabe 3

Bei dieser Aufgabe hat sich im Mathebuddy leider ein Fehler eingeschlichen. Statt \(\displaystyle-\frac{13}{9}\) sollte in dem Sammelsurium die Zahl \(\displaystyle-\frac{7}{5}\) auftauchen.

\(\displaystyle0,25=\frac{25}{100}=\frac{1}{4}\)

\(\displaystyle-1,4=-\frac{14}{10}=-\frac{7}{5}\)

\(\displaystyle3,6=\frac{36}{10}=\frac{18}{5}=3\frac{3}{5}\)

\(\displaystyle-1,5=-\frac{15}{10}=\frac{-3}{2}=\frac{-36}{24}\)

\(\displaystyle1,24=\frac{124}{100}=\frac{31}{25}\)

\(\displaystyle0,16=\frac{16}{100}=\frac{4}{25}\)

\(\displaystyle-2,2=-\frac{22}{10}=-\frac{11}{5}\)