ĐKXĐ: \(x>4\)
\(B=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{2024}{\sqrt{x}-2}\)
\(=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}-2}{2024}\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{2024}\)
\(=\dfrac{4}{2024\left(\sqrt{x}+2\right)}=\dfrac{1}{506\sqrt{x}+1012}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(B=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{2024}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{2024}\)
\(=\dfrac{4}{2024\left(\sqrt{x}+2\right)}=\dfrac{1}{506\left(\sqrt{x}+2\right)}\)
