A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)

ĐKXĐ : x > 0

ĐK:  \(x>0;x\ne1\)

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

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

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

\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(A>-1\) \(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}+1>0\)  \(\Leftrightarrow\)\(\frac{2\sqrt{x}-1}{\sqrt{x}}>0\)

Do  \(\sqrt{x}>0\)  \(\Rightarrow\)\(2\sqrt{x}-1>0\)\(\Leftrightarrow\)\(2\sqrt{x}>1\)\(\Leftrightarrow\)\(\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow\)\(x>\frac{1}{4}\)

Vậy  \(x>\frac{1}{4}\)\(\left(x\ne1\right)\)thì  A > - 1

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

Ta có: \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2-2\sqrt{x}+1}\)

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

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

Để \(A>-1\)thì \(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)\(\Leftrightarrow\sqrt{x}-1>-\sqrt{x}\)\(\Leftrightarrow2\sqrt{x}>1\)

\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)

Vậy \(A>-1\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\) 

Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)

               \(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

                \(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

\(a,ĐKXĐ:x\ge0;x\ne1\)

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

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

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

\(P=\left(x+1\right)^2\left(x-1\right)^2\)

\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)

\(P=\left(x^2+x-x-1\right)^2\)

\(P=\left(x^2-1\right)^2\)

b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)

\(\left(2-\sqrt{3}\right)^2\)

\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)

\(x^2-1< 2-\sqrt{3}\)

\(x^2< 3-\sqrt{3}\)

\(x< \sqrt{3-\sqrt{3}}\)

a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

b) Với x > = 0 và x khác 1

Ta có: \(P< 7-4\sqrt{3}\)

<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)

<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)

<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)

<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)

<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)

<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)

I don't know! :))