Aus Online Mathematik Brückenkurs 1

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Start by rewriting the terms on the left-hand side as one term having a common denominator {{Abgesetzte Formel||<math>\begin{align} \frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt] &= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt] &= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt] &= \frac{2}{(x-1)(x+1)}\,\textrm{.} \end{align}</math>}} If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as {{Abgesetzte Formel||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}} Because <math>x=1</math> cannot be a solution to the equation, the factor <math>x-1</math> can be removed from the denominator of both sides (i.e. actually, we multiply both sides by <math>x-1</math> and then eliminate it) {{Abgesetzte Formel||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}} Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators {{Abgesetzte Formel||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}} Expanding both sides {{Abgesetzte Formel||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}} The ''x''² terms cancel each other out and we obtain a first-order equation, {{Abgesetzte Formel||<math>3=5x-1\,,</math>}} which has the solution {{Abgesetzte Formel||<math>x=\frac{4}{5}\,\textrm{.}</math>}} We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation, {{Abgesetzte Formel||<math>\begin{align} \text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr) = \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl( \tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt] &= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25} = -\frac{50}{9}\cdot\frac{57}{50} = -\frac{57}{9} = -\frac{19}{3}\,,\\[15pt] \text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3} = \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}} = \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)} = \frac{19}{-3} = -\frac{19}{3}\,\textrm{.} \end{align}</math>}}

• Vorlage:Abgesetzte Formel

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