A1 (1-2, Romania) Solve the equations in the set of real numbers
(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) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=2$.
Solution 1 It is easy to know that $x \geqslant \frac{1}{2}$. Also,
$$
\begin{array}{c}
\sqrt{x+\sqrt{2 x-1}}=\frac{1}{\sqrt{2}}|\sqrt{2 x-1}+1|=\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1), \\
\sqrt{x-\sqrt{2 x-1}}=\frac{1}{\sqrt{2}}|\sqrt{2 x-1}-1| .
\end{array}
$$

It is easy to know that
$$
\begin{array}{c}
|\sqrt{2 x-1}-1|=1-\sqrt{2 x-1}, \frac{1}{2} \leqslant x \leqslant 1, \\
|\sqrt{2 x-1}-1|=\sqrt{2 x-1}-1,1 \leqslant x .
\end{array}
$$

Now let
$$
\begin{aligned}
y & =\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}} \\
& =\frac{1}{\sqrt{2}}(\sqrt{2 x-1}+1+|\sqrt{2 x-1}-1|),
\end{aligned}
$$

For $\frac{1}{2} \leqslant x \leqslant 1$, we have
$$
y=\sqrt{2} ;
$$

And for $1 \leqslant x$, we have
$$
y=\sqrt{2} \sqrt{2 x-1}(\geqslant \sqrt{2}) .
$$

Thus, the result of this problem is
(a) $\frac{1}{2} \leqslant x \leqslant 1$;
(b) no solution;
(c) $x=\frac{3}{2}$.

Solution 2 Let $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=y$, and square it, simplify, we get
$$
x+|x-1|=\frac{y^{2}}{2}, x \geqslant \frac{1}{2} \text {. }
$$

Or equivalently,
$$
\begin{array}{l}
1=\frac{y^{2}}{2}, \frac{1}{2} \leqslant x \leqslant 1, \\
2 x-1=\frac{y^{2}}{2}, 1 \leqslant x .
\end{array}
$$

Substitute $y=\sqrt{2}, 1,2$ into the above two equations respectively, we can get the results, the rest is the same as Solution 1.