\section*{Problem 6 - 071046}

Determine all real \(x\) that satisfy the following equation:

\[
\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2} \sqrt{\frac{x}{x+\sqrt{x}}}
\]
}

Obviously, \(x \geq \sqrt{x}\) for \(\sqrt{x-\sqrt{x}}\) to be defined. This is equivalent to \(x \geq 1\) or \(x=0\), but the latter is excluded because otherwise, due to \(x+\sqrt{x}=0\), a zero denominator would occur in the fraction under the square root on the right-hand side. Hence, from now on, let \(x \geq 1\).

By multiplying with \(\sqrt{x+\sqrt{x}} \neq 0\), the equation is equivalently transformed into

\[
x+\sqrt{x}-\sqrt{x^{2}-x}=\frac{3}{2} \sqrt{x} \quad \text{or} \quad x-\sqrt{x} \cdot \sqrt{x-1}=\frac{1}{2} \sqrt{x}
\]

After dividing by \(\sqrt{x} \neq 0\) and rearranging, we get \(\sqrt{x}-\frac{1}{2}=\sqrt{x-1}\). Squaring this equation, which is an equivalence transformation due to \(x \geq 1\) and thus \(\sqrt{x}-\frac{1}{2}>0\), leads to \(x-\sqrt{x}+\frac{1}{4}=x-1\) or \(\sqrt{x}=\frac{5}{4}\), hence \(x=\frac{25}{16}\).

Indeed, the (mathematically unnecessary) verification confirms that \(x=\frac{25}{16}\) is a solution to the original equation. As shown, this is also the only solution.