$3 \cdot 21 x$ takes what real value(s) when the equation
(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) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=2$.

all hold (here the square root takes non-negative values)?
[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 $x=\frac{3}{2}$.