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# Mathematrix: Aufgabenbeispiele/ Bruchtermegleichungen

Finden Sie die Definitions- und die Lösungsmenge der folgenden Bruchtermegleichung
${\displaystyle {\frac {2x-4}{x^{2}-4}}-{\frac {3x}{x^{2}-x}}={\frac {3x-13{,}5}{x^{2}-1}}}$

${\displaystyle \quad }$A) ${\displaystyle \quad {\frac {4x^{2}-5x-2x^{2}+4x}{x^{2}+x}}-{\frac {2x}{x-1}}={\frac {2x+15}{x^{2}-1}}\quad \Leftrightarrow }$

${\displaystyle \quad }$ ${\displaystyle {\frac {2x^{2}-x}{x^{2}+x}}-{\frac {2x}{x-1}}={\frac {2x+15}{x^{2}-1}}\quad \Leftrightarrow }$

${\displaystyle \quad }$ ${\displaystyle {\frac {x(2x-1)}{x(x+1)}}-{\frac {2x}{x-1}}={\frac {2x+15}{x^{2}-1}}\quad \Leftrightarrow }$

${\displaystyle \quad }$ ${\displaystyle {\frac {2x-1}{x+1}}-{\frac {2x}{x-1}}={\frac {2x+15}{(x-1)(x+1)}}\quad \Leftrightarrow }$

{\displaystyle {\begin{aligned}{\overset {\underbrace {\cdot \ (x\ -\ 1)} }{\frac {2x-1}{x+1}}}\qquad &\qquad -{\overset {\underbrace {\cdot \ (x\ +\ 1)} }{\frac {2x}{x-1}}}&&={\frac {2x+15}{(x-1)(x+1)}}\quad \Leftrightarrow \\\\{\frac {(2x-1)\cdot (x-1)}{(x+1)\cdot (x-1)}}&-{\frac {2x\cdot (x+1)}{(x-1)\cdot (x+1)}}&&={\frac {2x+15}{(x-1)(x+1)}}\quad \Leftrightarrow \\\\{\frac {2x^{2}-2x-x+1}{(x+1)\cdot (x-1)}}&-{\frac {2x^{2}+2x}{(x-1)\cdot (x+1)}}&&={\frac {2x+15}{(x-1)(x+1)}}\quad \Leftrightarrow \end{aligned}}}

${\displaystyle \quad }$${\displaystyle {\frac {2x^{2}-3x+1}{(x+1)\cdot (x-1)}}-{\frac {2x^{2}+2x}{(x-1)\cdot (x+1)}}={\frac {2x+15}{(x-1)(x+1)}}\qquad \qquad |\cdot (x-1)(x+1)\quad \Leftrightarrow }$

${\displaystyle \quad }$${\displaystyle \left[{\frac {2x^{2}-3x+1}{(x+1)\cdot (x-1)}}-{\frac {2x^{2}+2x}{(x-1)\cdot (x+1)}}\right]\cdot (x-1)(x+1)={\frac {2x+15}{(x-1)(x+1)}}\cdot (x-1)(x+1)\quad \Leftrightarrow }$

${\displaystyle \quad }$${\displaystyle {\frac {(2x^{2}-3x+1)\cdot {\xcancel {(x-1)(x+1)}}}{\xcancel {(x-1)(x+1)}}}-{\frac {(2x^{2}+2x)\cdot {\xcancel {(x-1)(x+1)}}}{\xcancel {(x-1)(x+1)}}}={\frac {(2x+15)\cdot {\xcancel {(x-1)(x+1)}}}{\xcancel {(x-1)(x+1)}}}\quad \Leftrightarrow }$

{\displaystyle {\begin{alignedat}{2}(2x^{2}-3x+1)&-(2x^{2}+2x)&&=2x+15\qquad |{\text{(Klammer auflösen)}}\quad \Leftrightarrow \\\\2x^{2}-3x+1&-2x^{2}-2x&&=2x+15\qquad |{\text{(Vereinfachen und Variablen trennen)}}\quad \Leftrightarrow \\\\&\quad -5x+1&&=2x+15\qquad |-1-2x)\quad \Leftrightarrow \\\\&\qquad -7x&&=\ 14\qquad \qquad |:(-7)\quad \Leftrightarrow \\\\&\qquad \quad \ x&&=-2\end{alignedat}}}

Die Definitionsmenge ist:

${\displaystyle x\neq \{-1,\ 0,\ 1\}\qquad }$ oder ${\displaystyle \mathbb {D} =\mathbb {R} \setminus \{-1,\ 0,\ 1\}}$

Die Lösungsmenge ist daher:

${\displaystyle \mathbb {L} =\{-2\}}$

${\displaystyle \quad }$B)