ĐKXĐ: \(x\ge0,x\ne1\)
\(M=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-\left(x+\sqrt{x}-2\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)
@Phạm Mỹ Dung Sao lúc nãy không gửi luôn?
Để \(M< -1\Leftrightarrow\frac{2}{x-1}< -1\)\(\Leftrightarrow\frac{2}{x-1}+1< 0\)
\(\Leftrightarrow\frac{2+x-1}{x-1}< 0\)\(\Leftrightarrow\frac{x+1}{x-1}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow-1< x< 1\)
(vì x < -1, x>1 không xảy ra)
Vậy để \(M< -1\Leftrightarrow-1< x< 1\)
