$2x^4-32$	=	0	| $+32$
$2x^4$	=	$32$	|:$2$
$x^4$	=	$16$	|
x1	=	$-\sqrt[4]{16}$	= $-2$
x2	=	$\sqrt[4]{16}$	= $2$

L={ $-2$; $2$}

D=R\{ $1$}

Wir multiplizieren den Nenner ${\left(x-1\right)}^2$ weg!

$\frac{16}{9}$	=	$\frac{1}{{\left(x-1\right)}^2}$	|⋅( ${\left(x-1\right)}^2$)
$\frac{16}{9}·{\left(x-1\right)}^2$	=	$\frac{1}{{\left(x-1\right)}^2}·{\left(x-1\right)}^2$
$\frac{16}{9}{\left(x-1\right)}^2$	=	$1$

$\frac{16}{9}{\left(x-1\right)}^2$	=	$1$	|⋅$\frac{9}{16}$
${\left(x-1\right)}^2$	=	$\frac{9}{16}$	|

1. Fall

2. Fall

(Alle Lösungen sind auch in der Definitionsmenge).

L={ $\frac{1}{4}$; $\frac{7}{4}$}