${\displaystyle \frac{-x}{x+3}}+2=1-{\displaystyle \frac{3x}{4\cdot (x-2)}}\backslash dfrac\{-x\}\{x+3\}+2=1-\backslash dfrac\{3x\}\{4\backslash cdot\; (x-2)\}$

Definitionsmenge bestimmen.

Im Nenner eines Bruches darf nicht Null stehen. Alle Zahlen, für die sich beim Einsetzen in die Gleichung im Nenner Null ergibt, dürfen nicht in der Definitionsmenge enthalten sein.

Setze jeden der Nenner gleich Null.

$x+3=0\Rightarrow x=-34x-8=0\Rightarrow x=2\backslash \; \backslash \; x+3=0\backslash qquad\backslash Rightarrow\backslash qquad\; x=-3\backslash \backslash 4x-8=0\backslash qquad\backslash Rightarrow\backslash qquad\; x=\backslash \; \backslash \; \backslash \; 2$

Der Nenner wird Null bei:

$x=-3\text{und}x=2\Rightarrow x=-3\backslash :\backslash text\{und\}\; \backslash :x=2\backslash qquad\backslash Rightarrow\backslash qquad$$D=\mathbb{R}\setminus \{-3;2\}D=\backslash mathbb\; R\backslash setminus\; \backslash left\backslash \{-3;2\backslash right\backslash \}$

Lösen der Bruchgleichung:

Bestimme den Hauptnenner, der HN ist: $4\cdot (x-2)\cdot (x+3)\backslash \; 4\backslash cdot(x-2)\backslash cdot(x+3)$

Multipliziere die Bruchgleichung mit dem HN:

${\displaystyle \frac{-x\cdot 4(x-2)\cancel{(x+3)}}{\cancel{(x+3)}}}+2\cdot 4(x-2)(x+3)=1\cdot 4(x-2)(x+3)-{\displaystyle \frac{3x\cdot \cancel{4(x-2)}(x+3)}{\cancel{4(x-2})}}\backslash dfrac\{\{-x\}\backslash cdot4(x-2)\backslash cancel\{(x+3)\}\}\{\backslash cancel\{(x+3)\}\}+2\backslash cdot4(x-2)(x+3)=1\backslash cdot4(x-2)(x+3)-\backslash dfrac\{3x\backslash cdot\backslash cancel\{4(x-2)\}(x+3)\}\{\backslash cancel\{4(x-2\})\}$

$-4x(x-2)+8(x-2)(x+3)=4(x-2)(x+3)-3x(x+3)\{-4x\}(x-2)+8(x-2)(x+3)=4(x-2)(x+3)-3x(x+3)$

$-4x^2+8x+8x^2+8x-48=4x^2+4x-24-3x^2-9x\backslash hspace\{4mm\}-4x^2+8x+8x^2+8x-48=4x^2+4x-24-3x^2-9x$

$4x^2+16x-48=x^2-5x-24\backslash hspace\{22.4mm\}4x^2+16x-48=x^2-5x-24$

$3x^2+21x-24=0\mathrm{\mid }:3\backslash hspace\{22.4mm\}3x^2+21x-24=0\backslash quad|:3$

$x^2+7x-8=0\backslash hspace\{27.4mm\}x^2+7x-8=0\backslash quad$

Löse die quadratische Gleichung mit der Mitternachtsformel:

$x_{1\mathrm{/}2}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\Rightarrow x_{1\mathrm{/}2}={\displaystyle \frac{-7\pm \sqrt{49+32}}{2}}x\_\{1/2\}=\backslash dfrac\{-b\backslash pm\backslash sqrt\{b^2-4ac\}\}\{2a\}\backslash qquad\backslash Rightarrow\backslash qquad\; x\_\{1/2\}=\backslash dfrac\{-7\backslash pm\backslash sqrt\{49+32\}\}\{2\}$

$x_{1\mathrm{/}2}={\displaystyle \frac{-7\pm 9}{2}}\Rightarrow x_1={\displaystyle \frac{-7-9}{2}}\text{und}{x}_2={\displaystyle \frac{-7+9}{2}}x\_\{1/2\}=\backslash dfrac\{-7\backslash pm9\}\{2\}\backslash qquad\backslash Rightarrow\backslash qquad\; x\_1=\backslash dfrac\{-7-9\}\{2\}\backslash qquad\; \backslash text\{und\}\backslash qquad\; x\_2=\backslash dfrac\{-7+9\}\{2\}$

$x_1=-8x_2=1x\_1=-8\backslash \backslash x\_2=\backslash \; \backslash \; \backslash \; 1$

Die Lösungsmenge:$\mathbb{L}=\{-8;1\}\backslash \; \backslash \; \backslash mathbb\{L\}=\backslash left\backslash \{-8;1\backslash right\backslash \}$