Lösung Bruchgleichungen
Version 1.1 von Stephanie Wietzorek am 2025/11/17 14:03
Löse unter Angabe der Definitionsmenge folgende Gleichungen:
- \(\frac{10}{x}=5 \)
\(D = \mathbb{R} \backslash \{0\}\)
\(L = \{2\}\)
2. \(\frac{10}{x+1}=5 \)
\(D = \mathbb{R} \backslash \{-1\}\)
\(L = \{1\}\)
3. \(\frac{10}{x+1}=\frac{5}{x-1} \)
\(D = \mathbb{R} \backslash \{-1;1\}\)
\(L = \{3\}\)
4. \(\frac{10}{x+1}=\frac{5x}{x-1}-\frac{5x^2}{x^2-1} \)
\(D = \mathbb{R} \backslash \{-1;1\}\)
\(\frac{10}{x+1}=\frac{5x}{x-1}-\frac{5x^2}{x^2-1} \) \(\quad HN =(x-1)(x+1)\)
\(=\frac{10(x-1)}{(x+1)(x-1)}=\frac{5x(x+1)}{(x-1)(x+1)}-\frac{5x^2}{x^2-1} |\cdot HN \)
\(=10(x-1)=5x(x+1)-5x^2 \)
\(=10x-10=5x^2+5x-5x^2 \)
\(=5x=10 \)
\(=x=2 \)
\(L = \{2\}\)
5. \(\frac{10}{2x+2}=\frac{5}{x+1}-1 \)
\(D = \mathbb{R} \backslash \{-1\}; \quad HN = 2(x+1)=2x+2\)
\(L = \{-1\}\)