\(A=\left[\dfrac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)+4\sqrt{x}+5}{\sqrt{x}+1}\\ A=\left[\dfrac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\cdot\dfrac{x-3\sqrt{x}-4+4\sqrt{x}+5}{\sqrt{x}+1}\\ A=\dfrac{1-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}+1}\)
\(A\ge2\Leftrightarrow\dfrac{-\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\ge0\Leftrightarrow\dfrac{-3\sqrt{x}-2}{\sqrt{x}+1}\ge0\)
Ta thấy \(\sqrt{x}+1\ge1>0;-3\sqrt{x}-2\le-2< 0\) nên \(A< 0,\forall x\)
Vậy ko có gt x nào thỏa mãn \(A\ge2\)
