\section*{Problem 1 - 081231}

Determine all real numbers \(x\) that satisfy the equation

\[
\frac{2 x}{a(x+a)}+\frac{1}{x-2 a}=\frac{4 x+6-a}{a(x+a)(x-2 a)}
\]

where \(a\) is a real number. (Case distinction!)
}

For the fractions to be defined, the denominators must be different from zero. Therefore, \(a \neq 0\), \(x \neq -a\), and \(x \neq 2a\). Under these conditions, the equation transforms through multiplication by the least common denominator into

\(2x(x-2a) + a(x+a) = 4x + 6 - a\) or

\[
0 = 2x^2 - 4ax + ax + a^2 - 4x - 6 + a = 2x^2 - (3a + 4)x + (a^2 + a - 6)
\]

This quadratic equation in \(x\) has the solutions

\[
\begin{gathered}
x_{1/2} = \frac{3a + 4}{4} \pm \sqrt{\frac{(3a + 4)^2}{16} - \frac{8 \cdot (a^2 + a - 6)}{16}} = \frac{3a + 4 \pm \sqrt{9a^2 + 24a + 16 - 8a^2 - 8a + 48}}{4} = \\
= \frac{3a + 4 \pm \sqrt{a^2 + 16a + 64}}{4} = \frac{3a + 4 \pm (a + 8)}{4}
\end{gathered}
\]

This results in

\[
x_1 = \frac{3a + 4 - (a + 8)}{4} = \frac{2a - 4}{4} = \frac{a}{2} - 1 \quad \text{and} \quad x_2 = \frac{3a + 4 + a + 8}{4} = a + 3
\]

It is necessary to exclude any extraneous solutions that do not satisfy the three conditions \(a \neq 0\), \(x \neq -a\), or \(x \neq 2a\).

1. Case: \(x_1 = -a\). Then \(\frac{a}{2} - 1 = -a\) or \(a - 2 = -2a\), so \(a = \frac{2}{3}\).

2. Case: \(x_2 = -a\). Then \(a + 3 = -a\), so \(a = -\frac{3}{2}\).

3. Case: \(x_1 = 2a\). Then \(\frac{a}{2} - 1 = 2a\) or \(a - 2 = 4a\), so \(a = -\frac{2}{3}\).

4. Case: \(x_2 = 2a\). Then \(a + 3 = 2a\), so \(a = 3\).

For all other real values of \(a\), the two values \(x_1 = \frac{a}{2} - 1\) and \(x_2 = a + 3\) are solutions to the original equation, which coincide exactly for \(a = -8\) and are otherwise distinct.