6"\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+4}\)\(=\dfrac{1}{\sqrt{x}-2}\)7:\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)=\dfrac{4}{x-4}\)Câu trả lời: 6"\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+4}\)\(=\dfrac{1}{\sqrt{x}-2}\)7:\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)=\dfrac{4}{x-4}\)