ĐKXĐ: \(x>4\)
\(A=\frac{\sqrt{\left(\sqrt{x-4}-2\right)^2}+\sqrt{\left(\sqrt{x-4}+2\right)^2}}{\sqrt{\left(1-\frac{4}{x}\right)^2}}=\frac{\left|\sqrt{x-4}-2\right|+\sqrt{x-4}+2}{1-\frac{4}{x}}\)
- Với \(x\ge8\Rightarrow A=\frac{2\sqrt{x-4}}{1-\frac{4}{x}}=\frac{2x\sqrt{x-4}}{x-4}\)
- Với \(4< x< 8\Rightarrow A=\frac{4}{1-\frac{4}{x}}=\frac{4x}{x-4}\)
b/
- Với \(x\ge8\Rightarrow A=\frac{2x}{\sqrt{x-4}}\)
Đặt \(\sqrt{x-4}=a\ge2\Rightarrow x=a^2+4\)
\(A=\frac{2\left(a^2+4\right)}{a}=2a+\frac{8}{a}\) nguyên \(\Rightarrow a=Ư\left(8\right)=\left\{2;4;8\right\}\)
\(\Rightarrow\sqrt{x-4}=\left\{2;4;8\right\}\Rightarrow x=\left\{8;20;68\right\}\)
- Với \(4< x< 8\Rightarrow A=\frac{4x}{x-4}=4+\frac{16}{x-4}\)
\(\Rightarrow x-4=Ư\left(16\right)\)
Mà \(4< x< 8\Rightarrow0< \sqrt{x-4}< 2\)
\(\Rightarrow\sqrt{x-4}=1\Rightarrow x=5\)
Vậy \(x=\left\{5;8;20;68\right\}\)
