$Đk:x>0;x\ne4$
$P=\dfrac{x+1}{\sqrt{x+2}-1}\\ =\dfrac{(x+1)(\sqrt{x+2}+1)}{x+2-1}=\sqrt{x+2}+1$
Khi $x=23$ thì $P=\sqrt{23+2}+1=6$
$Q=\bigg(\dfrac{3\sqrt{x}}{\sqrt{x}+2}-\dfrac{6x}{x-4}\bigg):\dfrac{\sqrt{x}}{2-\sqrt{x}}\\ =\dfrac{3\sqrt{x}(\sqrt{x}-2)-6x}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(-1)(\sqrt{x}-2)}{\sqrt{x}}\\ =\dfrac{\sqrt{x}(3\sqrt{x}-6-6\sqrt{x})(-1)(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{3(\sqrt{x}+2)}{\sqrt{x}+2}=3$
$P=Q\\ \Leftrightarrow \sqrt{x+2}+1=3\\ \Leftrightarrow\sqrt{x+2}=2\\ \Leftrightarrow x=2\ (t/m)$
Vậy $x=2$ là giá trị cần tìm