Đặt \(\sqrt{x}=a\)
\(A=\left(\dfrac{1}{a^2+a}-\dfrac{1}{a+1}\right):\dfrac{a-1}{a^2+2a+1}\)
\(=\left(\dfrac{a+1-a^2-a}{\left(a^2+a\right)\left(a+1\right)}\right).\dfrac{\left(a+1\right)^2}{a-1}\)
\(=\left[\dfrac{1-a^2}{a\left(a+1\right)^2}\right].\dfrac{\left(a+1\right)^2}{a-1}\)
\(=\dfrac{\left(1-a^2\right)\left(a+1\right)^2}{a\left(a+1\right)^2\left(a-1\right)}\)
\(=\dfrac{1-a^2}{a\left(a-1\right)}=\dfrac{1-a^2}{a^2-1}=\dfrac{\left(1-a\right)\left(1+a\right)}{\left(a-1\right)\left(a+1\right)}=\dfrac{1-a}{a-1}\)
\(\Rightarrow A=\dfrac{1-\sqrt{x}}{\sqrt{x}-1}\)
Vậy...
A = \(\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\)
A = \(\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)
A = \(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
A = \(\dfrac{-1\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
A = \(\dfrac{-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
