a) Với \(x\ge0;x\ne1\), ta có :
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(P=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)
\(P=[\frac{x-2\sqrt{x}+\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}].\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
Vậy : \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) Ta có : P > 0
\(\Leftrightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}-1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}< 1\end{cases}\Leftrightarrow}}\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)
Kết hợp với đk đề bài , ta được 0 < x < 1
Vậy với 0 < x < 1 thì P > 0
c) Với \(x=7-4\sqrt{3}=3-2.2.\sqrt{3}+4=\left(\sqrt{3}-2\right)^2\)thì :
\(P=-\sqrt{\left(\sqrt{3}-2\right)^2}\left(\sqrt{\left(\sqrt{3}-2\right)^2}-1\right)\)
\(P=-|\sqrt{3}-2|\left(|\sqrt{3}-2|-1\right)\)
\(P=\left(\sqrt{3}-2\right)\left(1-\sqrt{3}\right)\)
\(P=\sqrt{3}-3-3+2\sqrt{3}\)
\(P=3\sqrt{3}-5\)
Vậy với \(x=7-4\sqrt{3}\)thì \(P=3\sqrt{3}-5\)
d) Ta có \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Nhận thấy : \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu " = " xảy ra khi và chỉ khi
\(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì max P là \(\frac{1}{4}\)
