Coi như biểu thức xác định:
\(=\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}=x+\sqrt{x}+1\)
\(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}=x+\sqrt{x}+1\)