Lösung Potenzgleichungsmauer

Version 7.1 von Bastian Knöpfle am 2025/10/01 08:16

\(x=2\) \(x=\color{red}{-2}\)
\(x^5=\square\) \(x^\square=16\) \(x^\square+8=0\)
\(x=\square\) \(x=-3\)
\(x^\square=27\) \(x^{-2}=\square\) \(x^\square=-\frac{1}{3}\)

Hauptnavigation

	\(x=2\)		\(x=\color{red}{-2}\)
\(x^5=\square\)		\(x^\square=16\)		\(x^\square+8=0\)

	\(x=\square\)		\(x=-3\)
\(x^\square=27\)		\(x^{-2}=\square\)		\(x^\square=-\frac{1}{3}\)

unfertig	fertig
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