Đáp án:
b) x>9
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{2\sqrt x + 1 - \sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]\\
= \dfrac{{x - 2\sqrt x - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 3}}\\
= \dfrac{{ - 2\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 3}}\\
= - \dfrac{{2\sqrt x }}{{\sqrt x + 3}}\\
b)A > - 1\\
\to - \dfrac{{2\sqrt x }}{{\sqrt x + 3}} < - 1\\
\to \dfrac{{2\sqrt x }}{{\sqrt x + 3}} > 1\\
\to \dfrac{{2\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}} > 0\\
\to \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}} > 0\\
\to \sqrt x - 3 > 0\left( {do:\sqrt x + 3 > 0\forall x > 0} \right)\\
\to x > 9
\end{array}\)
