$2x^3$	=	$-250$	|:$2$
$x^3$	=	$-125$	|
$x$	=	$-\sqrt[3]{125}$	= $-5$

L={ $-5$}

D=R\{ $2$}

Wir multiplizieren den Nenner ${\left(x-2\right)}^2$ weg!

$\frac{16}{49}$	=	$\frac{1}{{\left(x-2\right)}^2}$	|⋅( ${\left(x-2\right)}^2$)
$\frac{16}{49}·{\left(x-2\right)}^2$	=	$\frac{1}{{\left(x-2\right)}^2}·{\left(x-2\right)}^2$
$\frac{16}{49}{\left(x-2\right)}^2$	=	$1$

$\frac{16}{49}{\left(x-2\right)}^2$	=	$1$	|⋅$\frac{49}{16}$
${\left(x-2\right)}^2$	=	$\frac{49}{16}$	|

1. Fall

2. Fall

(Alle Lösungen sind auch in der Definitionsmenge).

L={ $\frac{1}{4}$; $\frac{15}{4}$}