Đáp án:
x=25
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 9\\
B = \left[ {\dfrac{{x + 6\sqrt x + 9 - x + 6\sqrt x - 9 - 36}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right].\dfrac{{7\sqrt x - 2}}{{12}}\\
= \dfrac{{12\sqrt x - 36}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{7\sqrt x - 2}}{{12}}\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{7\sqrt x - 2}}{1}\\
= \dfrac{{7\sqrt x - 2}}{{\sqrt x + 3}}\\
P = A.B = \dfrac{{\sqrt x + 3}}{{2\sqrt x + 1}}.\dfrac{{7\sqrt x - 2}}{{\sqrt x + 3}}\\
= \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}}\\
\to 2P = \dfrac{{14\sqrt x - 4}}{{2\sqrt x + 1}} = \dfrac{{7\left( {2\sqrt x + 1} \right) - 11}}{{2\sqrt x + 1}}\\
= 7 - \dfrac{{11}}{{2\sqrt x + 1}}\\
De:P \in {Z^ + }\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{11}}{{2\sqrt x + 1}} < 7\\
2\sqrt x + 1 \in U\left( {11} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{11 - 14\sqrt x - 7}}{{2\sqrt x + 1}} < 0\\
\left[ \begin{array}{l}
2\sqrt x + 1 = 11\\
2\sqrt x + 1 = 1
\end{array} \right.
\end{array} \right.\left( {do:2\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
4 - 14\sqrt x < 0\\
\left[ \begin{array}{l}
2\sqrt x = 10\\
2\sqrt x = 0
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > \dfrac{4}{{49}}\\
\left[ \begin{array}{l}
x = 25\left( {TM} \right)\\
x = 0\left( l \right)
\end{array} \right.
\end{array} \right.
\end{array}\)
