điều kiện xác định : \(x\ge0\)
a) ta có : \(P=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\Leftrightarrow P=\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\Leftrightarrow P=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\dfrac{4\sqrt{x}}{3}\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
b) ta có : \(P=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\Leftrightarrow3Px-\left(3P+4\right)\sqrt{x}+3P=0\)
vì phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)
\(\Leftrightarrow\left(3P+4\right)^2-4\left(3P\right).\left(3P\right)\ge0\) \(\Leftrightarrow-27P^2+24P+16\ge0\)
\(\)\(\Leftrightarrow-27\left(x-\dfrac{4}{3}\right)\left(x+\dfrac{4}{9}\right)\ge0\Leftrightarrow-\dfrac{4}{9}\le x\le\dfrac{4}{3}\)
\(\Rightarrow P_{max}=\dfrac{4}{3}\) khi \(\sqrt{x}=\dfrac{-b}{2a}=\dfrac{3P+4}{6P}=\dfrac{3.\dfrac{4}{3}+4}{6.\dfrac{4}{3}}=1\) \(\Leftrightarrow x=1\)
\(\Rightarrow P_{min}=\dfrac{-4}{9}\) khi \(\sqrt{x}=\dfrac{-b}{2a}=\dfrac{3P+4}{6P}=\dfrac{3\left(\dfrac{-4}{9}\right)+4}{6\left(\dfrac{-4}{9}\right)}=-1\left(L\right)\)
ta nhận xét thấy \(P=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\ge0\)
\(\Rightarrow P_{min}=0\) khi \(x=0\)
vậy......................................................................................................................................
