Lời giải:
ĐKXĐ: $x\neq 4; x>2$
a)
\(A=\frac{\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x-1+2\sqrt{x-2}}}{\sqrt{x^2-8x+16}}.\frac{x-4}{x-2}\)
\(=\frac{\sqrt{(x-2)-2\sqrt{x-2}+1}+\sqrt{(x-2)+2\sqrt{x-2}+1}}{\sqrt{(x-4)^2}}.\frac{x-4}{x-2}\)
\(=\frac{\sqrt{(\sqrt{x-2}-1)^2}+\sqrt{(\sqrt{x-2}+1)^2}}{|x-4|}.\frac{x-4}{x-2}=\frac{|\sqrt{x-2}-1|+|\sqrt{x-2}+1|}{|x-4|}.\frac{x-4}{x-2}\)
Nếu $x>4$: \(A=\frac{\sqrt{x-2}-1+\sqrt{x-2}+1}{x-4}.\frac{x-4}{x-2}=\frac{2\sqrt{x-2}}{x-2}=\frac{2}{\sqrt{x-2}}\)
Nếu $3\leq x< 4$:
\(A=\frac{\sqrt{x-2}-1+\sqrt{x-2}+1}{4-x}.\frac{x-4}{x-2}=\frac{-2}{\sqrt{x-2}}\)
Nếu $2< x< 3$: \(A=\frac{1-\sqrt{x-2}+\sqrt{x-2}+1}{4-x}.\frac{x-4}{x-2}=\frac{2}{2-x}\)
b)
Nếu $x>4$: Để $A$ nguyên thì $2\vdots \sqrt{x-2}$
$\Rightarrow \sqrt{x-2}\in\left\{1;2\right\}$
$\Rightarrow x\in\left\{3; 6\right\}$. Vì $x>4$ nên $x=6$
Nếu $3\leq x< 4$: Để $A$ nguyên thì $-2\vdots \sqrt{x-2}$
$\Rightarrow \sqrt{x-2}\in\left\{1;2\right\}$
$\Rightarrow x=3$
Nếu $2< x< 3$: Không có giá trị nguyên nào nằm giữa $2;3$
