A double radical

Domain. We need $2x-1\ge0$, i.e. $x\ge\tfrac12$, and both radicals real, i.e. $x\pm\sqrt{2x-1}\ge0$. The condition $x\ge\sqrt{2x-1}$ is equivalent to $x^2\ge 2x-1$ (both sides nonnegative), i.e. $(x-1)^2\ge0$, which always holds. So the domain is $x\ge\tfrac12$.

Square. Set
$$a=\sqrt{x+\sqrt{2x-1}},\qquad b=\sqrt{x-\sqrt{2x-1}}.$$
Then
$$a^2+b^2=2x,\qquad ab=\sqrt{x^2-(2x-1)}=\sqrt{(x-1)^2}=|x-1|.$$
Since $a+b=2$, we get
$$4=(a+b)^2=a^2+b^2+2ab=2x+2|x-1|.$$
So $|x-1|=2-x$, which requires $x\le2$, and then splits into two cases.

Case $1\le x\le 2$: $|x-1|=x-1$, so
$$2x+2(x-1)=4\ \Longrightarrow\ 4x=6\ \Longrightarrow\ x=\tfrac32.$$
This lies in $[1,2]$.

Case $\tfrac12\le x<1$: $|x-1|=1-x$, so
$$2x+2(1-x)=4\ \Longrightarrow\ 2=4,$$
which is impossible.

Verification at $x=3/2$. $\sqrt{2x-1}=\sqrt2$, and
$$x+\sqrt{2x-1}=\frac32+\sqrt2=\frac{3+2\sqrt2}{2}=\frac{(1+\sqrt2)^2}{2},\qquad x-\sqrt{2x-1}=\frac32-\sqrt2=\frac{3-2\sqrt2}{2}=\frac{(1-\sqrt2)^2}{2}.$$
Hence
$$a=\frac{1+\sqrt2}{\sqrt2},\qquad b=\frac{\sqrt2-1}{\sqrt2},$$
and $a+b=\dfrac{1+\sqrt2+\sqrt2-1}{\sqrt2}=\dfrac{2\sqrt2}{\sqrt2}=2$. ✓

Conclusion. The equation has exactly one solution, $x=\dfrac32$.