Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
M = \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{x - 1}}\\
= \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 2} \right).\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right).\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {x + \sqrt x - 2} \right) - \left( {x - \sqrt x - 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
S = M.N = \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }} = \dfrac{2}{{\sqrt x - 1}}\\
c,\\
S < - 1\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 1}} < - 1\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 1}} + 1 < 0\\
\Leftrightarrow \dfrac{{2 + \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} < 0\\
\Leftrightarrow \sqrt x - 1 < 0\\
\Leftrightarrow 0 < x < 1
\end{array}\)
