a) Đkxđ: \(x\ge0\)
b) \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
c) Giả sử \(A\le1\Leftrightarrow\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\Leftrightarrow\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le1\)
\(\Leftrightarrow\sqrt{x}-x+\sqrt{x}-1\le0\Leftrightarrow-x+2\sqrt{x}-1\le0\Leftrightarrow-\left(\sqrt{x}-1\right)^2\le0\) (luôn đúng)
Vậy A \< 1 luôn đúng (đpcm)