\(P=\dfrac{x-1+9}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\)
\(\Leftrightarrow P>=2\cdot\sqrt{9}-2=2\cdot3-2=4\)
Dấu '=' xảy ra khi \(\sqrt{x}+1=3\)
hay x=4
\(\Leftrightarrow P\left(\sqrt{x}+1\right)=x+8\)
\(\Leftrightarrow x-P\sqrt{x}+8-P=0\) (*)
Coi pt (*) là pt bậc 2 ẩn \(\sqrt{x}\)
Pt (*) có nghiệm \(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\S\ge0\\P\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}P^2-4\left(8-P\right)\ge0\\P\ge0\\8-P\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}P\le-8;P\ge4\\P\ge0\\P\le8\end{matrix}\right.\)
\(\Rightarrow4\le P\le8\)
Vậy \(maxP=8\) (dấu bằng luôn xảy ra, muốn tìm x thay ngược vào pt (*))
