Đáp án:
x=25
Giải thích các bước giải:
\(\begin{array}{l}
c)B = \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) - 11\sqrt x + 11 - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 6\sqrt x - 11\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 5\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} = \dfrac{{\left( {2\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\sqrt x - 3}}{{\sqrt x + 3}}\\
P = A.B = \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}}.\dfrac{{2\sqrt x - 3}}{{\sqrt x + 3}}\\
= \dfrac{{2\sqrt x - 3}}{{\sqrt x + 2}} = \dfrac{{2\left( {\sqrt x + 2} \right) - 7}}{{\sqrt x + 2}}\\
= 2 - \dfrac{7}{{\sqrt x + 2}}\\
P \in Z \to \dfrac{7}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 7\\
\sqrt x + 2 = - 7\left( l \right)\\
\sqrt x + 2 = 1\\
\sqrt x + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 5\\
\sqrt x = - 1\left( l \right)\\
\sqrt x = - 3\left( l \right)
\end{array} \right.\\
\to x = 25
\end{array}\)
