$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\frac{\sqrt{x+4\sqrt{x-4}} +\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\frac{8}{x} +\frac{16}{x^{2}}}}\\ ĐK:\ x >4\\ A=\frac{\sqrt{x-4+4\sqrt{x-4} +4} +\sqrt{x-4-4\sqrt{x-4} +4}}{\sqrt{1-2.\frac{4}{x} +\left(\frac{4}{x}\right)^{2}}}\\ A=\frac{\sqrt{\left(\sqrt{x-4} +2\right)^{2}} +\sqrt{\left(\sqrt{x-4} -2\right)^{2}}}{\sqrt{\left( 1-\frac{4}{x}\right)^{2}}}\\ A=\frac{\sqrt{x-4} +2+|\sqrt{x-4} -2|}{1-\frac{4}{x}} \ ( x >4)\\ TH1:8 >x >4\\ A=\frac{\sqrt{x-4} +2-\sqrt{x-4} +2}{1-\frac{4}{x}} =\frac{4x}{x-4}\\ TH2:\ x\geqslant 8\\ A=\frac{\sqrt{x-4} +2+\sqrt{x-4} -2}{1-\frac{4}{x}} =\frac{2x\sqrt{x-4}}{x-4} =\frac{2x}{\sqrt{x-4}}\\ b.\ TH1:\ 8 >x >4\ mà\ x\in \mathbb{Z}\\ \Rightarrow x=\{7;6;5\} \ thử\ vào\ A\ thấy\ \ x=\{6;5\} \ thoả\ mãn\\ TH2:\ x\geqslant 8\\ Để\ \frac{2x}{\sqrt{x-4}} \in \mathbb{Z} \ đầu\ tiên\ x-4\ phải\ là\ số\ chính\ phương\ và\ 2x\vdots \sqrt{x-4}\\ \Rightarrow x=\{8\} \end{array}$
