a,
\(P=\left(\frac{2x}{x\sqrt{x}-x+\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(=\left[\frac{2x}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{x+1+\sqrt{x}}{x+1}\)
\(=\frac{2x-x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}.\frac{x+1}{x+\sqrt{x}+1}\)
\(=\frac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b,
\(P=-\frac{1}{7}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x+\sqrt{x}+1}=-\frac{1}{7}\)
\(\Leftrightarrow7\sqrt{x}+7=-x-\sqrt{x}-1\)
\(\Leftrightarrow x+8\sqrt{x}+8=0\)
\(\Leftrightarrow\left(\sqrt{x}+4\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+4=2\sqrt{2}\\\sqrt{x}+4=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\sqrt{2}-4\\\sqrt{x}=-2\sqrt{2}-4\end{matrix}\right.\)
\(\Rightarrow\text{phương trình vô nghiệm}\)
Vậy không tồn tại \(x\) thỏa mãn \(P=-\frac{1}{7}\)
