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