$Đk:x\ge0;x\ne4;x\ne9$
$a)$ Khi $x=64$ thì
$Q=\dfrac{\sqrt{64}-2}{\sqrt{64}-3}=\dfrac{8-2}{8-3}=\dfrac{6}{5}$
$b)P=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ =\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\ -đpcm-$
$c)K=Q(P-1)=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}.\bigg(\dfrac{\sqrt{x}}{\sqrt{x}-2}-1\bigg)\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}-3}.\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{2}{\sqrt{x}-3}$
Với $x\in Z$ thì $\dfrac{2}{\sqrt{x}-3}$ lớn nhất bằng 2 khi $\sqrt{x}-3=1\Leftrightarrow x=16$
Vậy $Max_K=2$ khi $x=16$
