Đáp án:
b. A<2
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} - 1} \right]:\left[ {\dfrac{{25 - x - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}} \right]\\
= \dfrac{{\sqrt x - \sqrt x - 5}}{{\sqrt x + 5}}:\dfrac{{25 - x - x + 9 + x - 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}{{ - x + 9}}\\
= \dfrac{5}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}{{x - 9}}\\
= \dfrac{5}{{\sqrt x + 3}}\\
b.A < 2\\
\to \dfrac{5}{{\sqrt x + 3}} < 2\\
\to \dfrac{{5 - 2\sqrt x - 6}}{{\sqrt x + 3}} < 0\\
\to \dfrac{{ - 1 - 2\sqrt x }}{{\sqrt x + 3}} < 0\\
\to - \dfrac{{1 + 2\sqrt x }}{{\sqrt x + 3}} < 0\\
\to \dfrac{{1 + 2\sqrt x }}{{\sqrt x + 3}} > 0\left( {ld} \right)\\
Do:x \ge 0;x \ne 9;x \ne 25\\
\to \left\{ \begin{array}{l}
1 + 2\sqrt x > 0\\
\sqrt x + 3 > 0
\end{array} \right.\\
\to A < 2
\end{array}\)
