Đáp án:
$M = {(\sqrt x - 1)^2}\left( {x \ge 0} \right)$
Giải thích các bước giải:
$\begin{array}{l}
M = \frac{{{x^2} - \sqrt x }}{{x + \sqrt x + 1}} - \frac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} + x + 1\\
Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x + \sqrt x + 1 \ne 0\\
x - \sqrt x + 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge 0\\
{\left( {\sqrt x + \frac{1}{2}} \right)^2} + \frac{3}{4} \ne 0)luon\,dung\\
{\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ne 0)luon\,dung
\end{array} \right.\\
\Rightarrow x \ge 0\\
M = \frac{{\sqrt x \left( {x\sqrt x - 1} \right)}}{{x + \sqrt x + 1}} - \frac{{\sqrt x \left( {x\sqrt x + 1} \right)}}{{x - \sqrt x + 1}} + x + 1\\
M = \frac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}} - \frac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} + x + 1\\
M = \sqrt x \left( {\sqrt x - 1} \right) - \sqrt x \left( {\sqrt x + 1} \right) + x + 1\\
M = x - \sqrt x - x - \sqrt x + x + 1\\
M = x - 2\sqrt x + 1\\
M = {(\sqrt x - 1)^2}\left( {x \ge 0} \right)
\end{array}$
