a) Ta có: \(M=\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{\sqrt{x}+1}{x}\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\frac{\sqrt{x}+1}{x}\)
\(=\left(\frac{x-1}{\sqrt{x}}\right):\frac{\sqrt{x}+1}{x}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{x}{\sqrt{x}+1}\)
\(=\sqrt{x}\cdot\left(\sqrt{x}-1\right)\)
b) Để M<0 thì \(\sqrt{x}\cdot\left(\sqrt{x}-1\right)< 0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
Vậy: Để M<0 thì 0<x<1
