Đặt \(\sqrt{x}=a\)
\(V=\left(\dfrac{1}{a+2}+\dfrac{1}{x-2}\right)\dfrac{a+2}{a}\)
\(=\left(\dfrac{a-2+a+2}{\left(a+2\right)\left(a-2\right)}\right)\dfrac{a+2}{a}\)
\(=\dfrac{2a\left(a+2\right)}{\left(a+2\right)\left(a-2\right)a}\)
\(=\dfrac{2}{a-2}\)
\(\Rightarrow V=\dfrac{2}{\sqrt{x}-2}\)