Bulgarian Spring Tournament

Note that $x\ne 4$ and $|x-1|=x-1$ for $x>1$, otherwise $|x-1|=1-x$.

If $x>1$ we get: $$\frac{x^2-8x+7}{x-4}\le 0⟺\frac{(x-7)(x-1)}{x-4}\le 0⟺x\in (4,7]$$

If $x\le 1$ we get: $$\frac{x^2-10x+9}{x-4}\le 0\iff \frac{(x-9)(x-1)}{x-4}\le 0\iff x\in (-\infty ,1]$$

Finally $x\in (-\infty ,1]\cup (4,7]$. $\square$