Đáp án:
\(\begin{array}{l}
b)B = \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c)MaxP = 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)B = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2 + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c)P = A:B = \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}:\dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x }} = 1 + \dfrac{3}{{\sqrt x }}\\
P\max \Leftrightarrow \dfrac{3}{{\sqrt x }}\max \\
\Leftrightarrow \sqrt x \min \\
\to \sqrt x = 1\\
\to x = 1\\
\to MaxP = 1 + \dfrac{3}{{\sqrt 1 }} = 4
\end{array}\)
