Đáp án:
b. \(\left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)P = \dfrac{{2x + 4\sqrt x + 6 + \left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right) + 3\left( {\sqrt x - 1} \right) - 2\left( {x + 2\sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 4\sqrt x + 6 + x + \sqrt x - 6 + 3\sqrt x - 3 - 2x - 4\sqrt x + 6}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 4\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
2)P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}} = 1 + \dfrac{2}{{\sqrt x - 1}}\\
P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 2\\
\sqrt x - 1 = - 2\left( l \right)\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
