Đáp án:
\(\left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 1\\
B = \left[ {\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}} \right].\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - x + \sqrt x + 2}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{x - 1}}\\
B \in Z \Leftrightarrow \dfrac{2}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 = - 2\left( l \right)\\
x - 1 = 1\\
x - 1 = - 1\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.
\end{array}\)
