$A= \dfrac{7 \sqrt{7}-8}{11+2 \sqrt{7}} - \sqrt{(1- \sqrt{7})^2} \\
= \dfrac{(7 \sqrt{7}-8)(11-2 \sqrt{7})}{11^2-(2 \sqrt{7})^2} - \left |1- \sqrt{7} \right |$
$=\dfrac{77 \sqrt{7}-14.7-88+16 \sqrt{7}}{121-28} - (\sqrt{7}-1)$ (vì $1< \sqrt{7}$)
$=\dfrac{77 \sqrt{7}-14.7-88+16 \sqrt{7}}{93} - \sqrt{7}+1 \\
= \dfrac{93 \sqrt{7}-186}{93} - \sqrt{7}+1 \\
=\dfrac{93 (\sqrt{7}-2)}{93} - \sqrt{7}+1 \\
= \sqrt{7}-2 -\sqrt{7}+1=-1$
Vậy $A=-1$