algebra problem, Solve the equation 
a nice problem: Solve the equation $$\left|2x-57-2\sqrt{x-55}+\frac{1}{x-54-2\sqrt{x-55}}\right|=|x-1|.$$

It's just for sharing a new ideas, thanks:)
 A: Now things become easy. Rewriting the LHS in the form
$\begin{align*} 
&(x-1) + (\sqrt {x-55}-1)^2+\frac{1}{ (\sqrt {x-55}-1)^2}-2\\
&= (x-1)\\
&\;\; +\left((\sqrt {x-55}-1)-\frac{1}{(\sqrt {x-55}-1)}\right)^2 \end{align*}$
So the solution is $x=59$ and $x=55$.
A: The left hand side of expression, by completing the square, is:
\begin{align*}
& \left|(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1 + x - 3 + \frac{1}{(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1}\right| \\
= & \left|(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} + x - 3\right|.
\end{align*}
Since $x \geq 55$, the expression inside the absolute symbol is positive, so is the one in the right hand side, so the equation can be written as
\begin{align*}
(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} + x - 3 & = x - 1 \\
(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} & = 2 \tag{1}
\end{align*}
But by the algebraic-geometric inequality, the left hand side of $(1)$ is at least $2$, and the equality holds if and only if 
$$(\sqrt{x - 55} - 1)^2 = 1$$
which gives $x = 55$ and $x = 59$.
