a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+\sqrt{x}+1}{x+1}\)
\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
b) Để \(P\le0\) thì \(-\left(\sqrt{x}-1\right)\le0\)
\(\Leftrightarrow\sqrt{x}\ge1\)
hay \(x\ge1\)
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