\(A=\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x-\sqrt{x}}:\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x-1}}\right)\left(x>0;x\ne1\right)\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x-1}\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x-1}-\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-1}\right)\)
\(=\frac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}:\left(\frac{\left(x-1\right)\left(\sqrt{x}-2\right)}{x-1}\right)\)
\(=\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}-2}=\frac{1}{x-2\sqrt{x}}\)
Để biểu thức của A có giá trị âm
\(\Leftrightarrow\frac{1}{x-2\sqrt{x}}>0\)
\(\Leftrightarrow x-2\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}KTM\\x>4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\)
\(\Rightarrow0< x< 4\)
Vậy để giá trị của A có giá trị âm thì 0<x<4
2)\(\sqrt{9x-9}+1=13\) ĐKXĐ:\(x\ge1\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=2\)
\(\Leftrightarrow x=3\)
Vậy x=3 để pt thỏa mãn
