Solve the following equation:

$$
\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2} \sqrt{\frac{x}{x+\sqrt{x}}}
$$
For the expressions in the equation to make sense, it is necessary that $x>0$, and further, $x-\sqrt{x} \geq 0$ must hold. The latter is true for positive $x$ if and only if

$$
x \geq 1
$$

For such $x$, we can divide (2) by the right-hand side square root expression, and the equation holds for $x$ not less than 1 if and only if

$$
\frac{x+\sqrt{x}}{\sqrt{x}}+\sqrt{\frac{x^{2}-x}{x}}=\sqrt{x}+1-\sqrt{x-1}=\frac{3}{2}
$$

which is true if and only if

$$
\sqrt{x}-\frac{1}{2}=\sqrt{x-1}
$$

Here, the left side is positive under the condition (3), so the two sides (with the roots being non-negative) are equal if and only if their squares are equal:

$$
x-\sqrt{x}+\frac{1}{4}=x-1, \quad \sqrt{x}=\frac{5}{4}, \quad x=\frac{25}{16} .
$$

This is a number greater than 1, so it is the only solution to the equation.

Remark: Writing (4) in the form $\sqrt{x}-\sqrt{x-1}=1 / 2$ and then multiplying by the positive $2(\sqrt{x}+\sqrt{x-1})$ expression, the equation $\sqrt{x}+\sqrt{x-1}=2$ is obtained, from which and the previous one, as a system of linear equations, $\sqrt{x}=5 / 4$ (and the consistent result $\sqrt{x-1}=3 / 4$) is obtained again.