\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?