3. 21 For what real values of $x$ do the equations
(a) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$,
(b) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=1$,
(c) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=2$.

hold (where the square roots are taken to be non-negative)?
[Solution] Construct the auxiliary function
$$
f(x)=\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}},
$$

then the domain of $f(x)$ is $\left\{x \left\lvert\, x \geqslant \frac{1}{2}\right.\right\}$.
Notice that $(\sqrt{2 x-1}+1)^{2}=2(x+\sqrt{2 x-1})$
and $(\sqrt{2 x-1}-1)^{2}=2(x-\sqrt{2 x-1})$,
so we have $\sqrt{x+\sqrt{2 x-1}}=\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1)$
and $\sqrt{x-\sqrt{2 x-1}}=\frac{1}{\sqrt{2}}|\sqrt{2 x-1}-1|$,
thus the function $f(x)$ is
$$
f(x)=\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1+|\sqrt{2 x-1}-1|)
$$

Also, when $\sqrt{2 x-1}-1>0$, i.e., $x>1$,
$$
|\sqrt{2 x-1}-1|=\sqrt{2 x-1}-1
$$

When $\sqrt{2 x-1}-1 \leqslant 0$, i.e., $\frac{1}{2} \leqslant x \leqslant 1$,
$$
|\sqrt{2 x-1}-1|=1-\sqrt{2 x-1}
$$

Below, we discuss $f(x)$ in two cases,
(1) When $\frac{1}{2} \leqslant x \leqslant 1$,
$$
\begin{aligned}
f(x) & =\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1+1-\sqrt{2 x-1}) \\
& =\sqrt{2}
\end{aligned}
$$
(2) When $x>1$,
$$
\begin{aligned}
f(x) & =\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1+\sqrt{2 x-1}-1) \\
& =\sqrt{2} \cdot \sqrt{2 x-1}>\sqrt{2} .
\end{aligned}
$$

For equation (a), by (1), its solution is
$$
\frac{1}{2} \leqslant x \leqslant 1
$$

For equation (b), by (1) and (2), within the domain,
$$
f(x) \geqslant \sqrt{2} .
$$

Thus, $(b) f(x)=1$ has no solution.
For equation (c), we solve the equation
$$
\begin{array}{l}
\sqrt{2} \cdot \sqrt{2 x-1}=2 \\
x=\frac{3}{2}>1 .
\end{array}
$$

So its solution is $x=\frac{3}{2}$.