Suppose you have two expressions $e_1$ and $e_2$ and you know
$$e_1 = e_2.$$
Then, if you apply a function to both sides, you have
$$f(e_1) = f(e_2).$$
However, this logic in general does not reverse, unless the function
$f$ is 1-1. This is the mechanism by which extraneous roots get introduced.
When you square both sides of an equation, you are destroying information about the signs of the two sides. Now, the equality will match if the two sides have the same absolute value. This process can, and often does, introduce spurious roots.
Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.
As a very simple example, notice the following two equations:
$$x = \sqrt 4 \iff x = +2$$
$$x^2 = 4 \iff \vert x\vert = 2 \iff x = \pm 2$$
The first equation has only one solution: $+\sqrt 4$. The second, however, has two solutions: $\pm\sqrt 4$. And you get the second equation by squaring the first one.
The exact same idea applies to your example. You have
$$\sqrt{2x-3} = 3-x$$
which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving
$$0 = 12-8x+x^2 \iff \color{blue}{\pm}\sqrt{2x-3} = 3-x$$
which has a $\pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:
$$2x-3 = 9-6x+x^2; \quad \color{blue}{x \leq 3}$$
$$0 = 12-8x+x^2; \quad \color{blue}{x \leq 3}$$
Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.
Best Answer
It's $$\sum_{cyc}\left(x+\frac{1}{x}+2-2\sqrt{2x+1}\right)=0$$ or $$\sum_{cyc}\frac{x^2-2x\sqrt{2x+1}+2x+1}{x}=0$$ or $$\sum_{cyc}\frac{(x-\sqrt{2x+1})^2}{x}=0,$$ which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain: $$x=\sqrt{2x+1}$$ and $$y=\sqrt{2y+1},$$ which gives $$x=y=1+\sqrt2.$$