Absolutely Absolute

$\begin{aligned} \left||x^2-5x+6| - |2x-4|\right| & = \left|x^2-3x+2\right| \\ \left||(x-2)(x-3)| - 2|x-2|\right| & = \left|(x-2)(x-1)\right| \\ \left||x-3| - 2\right| & = \left|x-1\right| \end{aligned}$

Let us assume that the maximum value of $x$ satisfying the equation is $> 0$ .

For $0 < x \le 1$ , $\begin{cases} ||x-3|-2| = |3-x-2| = 1-x \\ |x-1| = 1-x \end{cases}$

Therefore, $LHS = RHS$ $\implies x \in (0, 1]$ satisfies the equation.

For $1 < x \le 3$ , $\begin{cases} ||x-3|-2| = |3-x-2| = x-1 \\ |x-1| = x-1 \end{cases}$

Again $LHS = RHS$ $\implies x \in (0, 3]$ satisfies the equation.

For $x > 3$ , $\begin{cases} ||x-3|-2| = |x-3-2| = \begin{cases} 5-x & \text{ if }x \le 5 \\ x-5 & \text{ if }x > 5 \end{cases} \\ |x-1| = x -1 \end{cases}$

Therefore, $LHS \ne RHS$ for $x > 3$ . And the maximum value of $x$ satisfying the equation is $\boxed{3}$ .