Đáp án:
b) x>1
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)M = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\left[ {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\dfrac{{x + 1 - 2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}.\dfrac{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 1}}\\
b)M > 1\\
\to \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 1}} > 1\\
\to \dfrac{{x + \sqrt x + 1 - \sqrt x + 1}}{{\sqrt x - 1}} > 0\\
\to \dfrac{{x + 2}}{{\sqrt x - 1}} > 0\\
\to \sqrt x - 1 > 0\left( {do:x + 2 > 0\forall x \ge 0} \right)\\
\to x > 1
\end{array}\)
