a) \(P=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{2x}{x-4}=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\dfrac{x-4}{2x}=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2x}=\dfrac{2\sqrt{x}}{2x}=\dfrac{1}{\sqrt{x}}\)
b) Ta có \(P< 1\Leftrightarrow\dfrac{1}{\sqrt{x}}< 1\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Kết hợp với ĐK: vậy x>1 và \(x\ne4\) thì P<1