ĐKXĐ: ...
\(M=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x}{x+\sqrt{x}+1}\)
\(M< \frac{1}{3}\Leftrightarrow\frac{x}{x+\sqrt{x}+1}-\frac{1}{3}< 0\Leftrightarrow\frac{2x-\sqrt{x}-1}{x+\sqrt{x}+1}< 0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{x+\sqrt{x}+1}< 0\Rightarrow\sqrt{x}-1< 0\Rightarrow x< 1\)
Đề bài sai, \(M< \frac{1}{3}\) khi \(x< 1\) còn \(x>1\) thì \(M>\frac{1}{3}\)