đk x ≠ 1; x ≥ 0
\(A=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(x-1\right)}\\ =\left(\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(x-1\right)}\right).\dfrac{\sqrt{x}\left(x-1\right)}{\sqrt{x}^3-1}\\ =\dfrac{1}{\sqrt{x}-1}\)