\ANG{
  Löse jeweils die Klammer auf: \\ \\
  Beispiel: \quad $\D \left( \frac p {q^2}\right)^{-1} = \frac 1 {\frac p {q^2}} = 1 \, : \, \frac p {q^2} = 1 \cdot \frac {q^2} p = \frac {q^2} p$ \\ \\
  (1) \quad $\D \left( \frac {p^2} q \right)^{-1}$
  \qquad (2) \quad $\D \left( \frac p {q^2} \right)^{-2}$
  \qquad (3) \quad $\D \left( \frac p {q^{-1}} \right)^2$
}
%
\LSG{
(1) \quad $\D \left( \frac {p^2} q \right)^{-1} = \frac 1 {\frac {p^2} q} = 1 \, : \, \frac {p^2} q = 1 \cdot \frac q {p^2} = \frac q {p^2}$ \\ \\
(2) \quad $\D \left( \frac p {q^2}\right)^{-2} = \frac 1 {\left( \frac p {q^2} \right)^2} = \frac 1 { \frac {p^2} {q^4} } = 1 \, : \, \frac {p^2} {q^4} = 1 \cdot \frac {q^4} {p^2} = \frac {q^4} {p^2}$ \\ \\
(3) \quad $\D \left( \frac p {q^{-1}} \right)^2 = \left( \frac p {\frac 1 q} \right)^2
= \left( p : \frac 1 q \right)^2 =  \left( p \cdot \frac q 1 \right)^2 = \left( p \cdot q \right)^2 = p^2 \, q^2$ \\ \\
oder \quad $\D \left( \frac p {q^{-1}} \right)^2 = \frac {p^2}{q^{-2}} = \frac {p^2}  {\frac 1 {q^2}} = p^2 \cdot \frac {q^2} 1 = p^2 \, q^2$
}