4) \(P=\dfrac{2}{\sqrt{3}+1}-\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{4}{\sqrt{5}+1}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)}{2}-\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}+\dfrac{4\left(\sqrt{5}-1\right)}{4}\)
\(=\sqrt{3}-1-\sqrt{5}-\sqrt{3}+\sqrt{5}-1=-2\)
5) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)
Để \(A\in Z\Rightarrow5⋮\sqrt{x}+3\) mà \(\sqrt{x}+3\ge3\Rightarrow\sqrt{x}+3=5\Rightarrow x=4\)