(\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\)).(\(\frac{1}{\sqrt{x}+1}\)+\(\frac{1}{x-1}\)) (Điều kiện xác định: x ≠ 1; x≥0)
=(\(\frac{x}{x+\sqrt{x}}-\frac{1}{x+\sqrt{x}}\)).(\(\frac{\sqrt{x}-1}{x-1}\)+\(\frac{1}{x-1}\))
= \(\frac{x-1}{x+\sqrt{x}}\).\(\frac{\sqrt{x}-1+1}{x-1}\)
=\(\frac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\).\(\frac{\sqrt{x}}{x-1}\)
=\(\frac{1}{\sqrt{x}+1}\)