điều kiện xác định : \(x\ge0\)
ta có : \(P=\dfrac{x^2+x+1}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1+x^2-\sqrt{x}}{x+\sqrt{x}+1}\)
\(=1+\dfrac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}=1+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(=1+\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow GTNN\) của \(P\) là \(\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
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