Đáp án:
\(\begin{array}{l}
a)A = \dfrac{{3\sqrt x }}{{\sqrt x + 3}}\\
b)P = \dfrac{3}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{7\sqrt x + 3}}{{9 - x}}x + \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 7\sqrt x - 3 + 2x - 6\sqrt x + x + 4\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x - 9\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{{3\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 3}}\\
b)\,A.B = \dfrac{{3\sqrt x }}{{\sqrt x + 3}}.\dfrac{{\sqrt x + 7}}{{3\sqrt x }} = \dfrac{{\sqrt x + 7}}{{\sqrt x + 3}}\\
Thay\,x = 25\,ta\,co\\
P = \dfrac{{\sqrt {25} + 7}}{{\sqrt {25} + 3}} = \dfrac{{12}}{8} = \dfrac{3}{2}
\end{array}\)
