\(ĐKXD:x\ge0;x\ne1\)
\(A=\left(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)=\left(\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)=\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)\)
\(A=\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)=\frac{2\sqrt{x}-x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1}{x+1}=\frac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1}{x+1}=\frac{-\sqrt{x}+1}{x+1}:\frac{x+\sqrt{x}+1}{x+1}=\frac{\left(-\sqrt{x}+1\right)\left(x+1\right)}{\left(x+1\right)\left(x+\sqrt{x}+1\right)}=\frac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(A\le0\Leftrightarrow-\sqrt{x}+1\le0\left(vì:x+\sqrt{x}+1>0\right)\Leftrightarrow-\sqrt{x}\le-1\Leftrightarrow\sqrt{x}\ge1\Leftrightarrow x\ge1.Vậy:x\ge1thi:A\ge0\)
