Đáp án:
a) \(\dfrac{{3\sqrt x + 8}}{{3\left( {\sqrt x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
B = \dfrac{{3\sqrt x \left( {\sqrt x + 1} \right) + 2\sqrt x - 8}}{{3\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{3x + 3\sqrt x + 2\sqrt x - 8}}{{3\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{3x + 5\sqrt x - 8}}{{3\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 8} \right)}}{{3\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{3\sqrt x + 8}}{{3\left( {\sqrt x + 1} \right)}}\\
b)P = A.B = \dfrac{7}{{3\sqrt x + 8}}.\dfrac{{3\sqrt x + 8}}{{3\left( {\sqrt x + 1} \right)}}\\
= \dfrac{7}{{3\sqrt x + 3}}\\
P \in Z \to \dfrac{7}{{3\sqrt x + 3}} \in Z\\
\to 3\sqrt x + 3 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
3\sqrt x + 3 = 7\\
3\sqrt x + 3 = 1\left( l \right)
\end{array} \right.\\
\to \sqrt x = \dfrac{4}{3}\\
\to x = \dfrac{{16}}{9}
\end{array}\)
