Mathematrix: Aufgabenbeispiele/ Bruchtermegleichungen

{{#lst:Mathematrix: Aufgabensammlung/_Bruchtermegleichungen|01}}

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

$$ $\frac{2x^2-x}{x^2+x}-\frac{2x}{x-1}=\frac{2x+15}{x^2-1}\iff$

$$ $\frac{x(2x-1)}{x(x+1)}-\frac{2x}{x-1}=\frac{2x+15}{x^2-1}\iff$

$$ $\frac{2x-1}{x+1}-\frac{2x}{x-1}=\frac{2x+15}{(x-1)(x+1)}\iff$

$\begin{array}{cccc}\stackrel{\underbrace{\cdot (x-1)}}{\frac{2x-1}{x+1}}& -\stackrel{\underbrace{\cdot (x+1)}}{\frac{2x}{x-1}}& & =\frac{2x+15}{(x-1)(x+1)}\iff \\ \\ \frac{(2x-1)\cdot (x-1)}{(x+1)\cdot (x-1)}& -\frac{2x\cdot (x+1)}{(x-1)\cdot (x+1)}& & =\frac{2x+15}{(x-1)(x+1)}\iff \\ \\ \frac{2x^2-2x-x+1}{(x+1)\cdot (x-1)}& -\frac{2x^2+2x}{(x-1)\cdot (x+1)}& & =\frac{2x+15}{(x-1)(x+1)}\iff \end{array}$

$$$\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)}|\cdot (x-1)(x+1)\iff$

$$$[\frac{2x^2-3x+1}{(x+1)\cdot (x-1)}-\frac{2x^2+2x}{(x-1)\cdot (x+1)}]\cdot (x-1)(x+1)=\frac{2x+15}{(x-1)(x+1)}\cdot (x-1)(x+1)\iff$

$$$\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)}}\iff$

$\begin{array}{cccc}(2x^2-3x+1)& -(2x^2+2x)& & =2x+15|\text{(Klammer auflösen)}\iff \\ \\ 2x^2-3x+1& -2x^2-2x& & =2x+15|\text{(Vereinfachen und Variablen trennen)}\iff \\ \\ & -5x+1& & =2x+15|-1-2x)\iff \\ \\ & -7x& & =14|:(-7)\iff \\ \\ & x& & =-2\end{array}$

Die Definitionsmenge ist:

$x\ne \{-1,0,1\}$ oder $𝔻=ℝ\setminus \{-1,0,1\}$

Die Lösungsmenge ist daher:

$𝕃=\{-2\}$

$$B)