\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{\sqrt{x}+2}\)
\(ĐK:x\ge0;x\ne1\)
Biểu thức trở thành:
\(\left(\dfrac{2\sqrt{x}+x}{\sqrt{x}^3-1}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}^3-1}\right).\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\sqrt{x}^3-1}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{1}{\sqrt{x}+2}\)
