Đáp án:
\(\begin{array}{l}
a)\dfrac{{10}}{3}\\
b)B = \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c)x\max = 24
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 25\\
\to A = \dfrac{{\sqrt {25} + 5}}{{\sqrt {25} - 2}} = \dfrac{{10}}{3}\\
b)B = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2 + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c)P = B:A = \dfrac{{\sqrt x }}{{\sqrt x - 2}}:\dfrac{{\sqrt x + 5}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 5}}\\
P < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x }}{{\sqrt x + 5}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - \sqrt x - 5}}{{2\left( {\sqrt x + 5} \right)}} < 0\\
\to \dfrac{{\sqrt x - 5}}{{2\left( {\sqrt x + 5} \right)}} < 0\\
\to \sqrt x - 5 < 0\left( {do:\sqrt x + 5 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 25;x \ne 4\\
\to x\max = 24
\end{array}\)
