a)  \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(P=\left(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)

\(\Leftrightarrow P=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)\)

\(\Leftrightarrow P=\frac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\frac{x+1}{x+\sqrt{x}+1}\)

\(\Leftrightarrow P=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\frac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)

b) Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Để \(P\le0\Leftrightarrow-\sqrt{x}+1\le0\)

\(\Leftrightarrow-\sqrt{x}\le-1\)

\(\Leftrightarrow\sqrt{x}\ge1\)

\(\Leftrightarrow x\ge1\)

Vì đkxđ : \(x\ne1\)

Vậy để \(P\le0\Leftrightarrow x>1\)

\(ĐKXĐ:0\le x\ne x\)

a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)

\(P=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)

\(P=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}.\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow MAX_P=\frac{1}{4}\text{ khi }x=\frac{1}{4}\)

a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2 \)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:

A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)

a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)

a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)

b) Ta có   \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)

\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)

Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)