với x\(\ge\)0 và \(x\ne1\), ta có:
\(A=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x+1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(x+1\right)}{\sqrt{x}+1}\)