Đáp án:
a) \(\dfrac{{ - x + 4}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - \dfrac{1}{2};0;\dfrac{1}{2};2} \right\}\\
B = \dfrac{x}{2} - \dfrac{{1 - 2x - x\left( {4x - 2} \right)}}{{x\left( {1 - 2x} \right)}}:\dfrac{{{x^2} - 2x + 4x - {x^2} + 1}}{{x\left( {x - 2} \right)}}\\
= \dfrac{x}{2} - \dfrac{{1 - 2x - 4{x^2} + 2x}}{{x\left( {1 - 2x} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{2x + 1}}\\
= \dfrac{x}{2} - \dfrac{{1 - 4{x^2}}}{{x\left( {1 - 2x} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{2x + 1}}\\
= \dfrac{x}{2} - \dfrac{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}{{x\left( {1 - 2x} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{2x + 1}}\\
= \dfrac{x}{2} - \left( {x - 2} \right)\\
= \dfrac{{x - 2x + 4}}{2} = \dfrac{{ - x + 4}}{2}\\
b)B = 1\\
\to \dfrac{{ - x + 4}}{2} = 1\\
\to 4 - x = 2\\
\to x = 2\\
c)B \in Z\\
\Leftrightarrow \dfrac{{ - x + 4}}{2} \in Z\\
\Leftrightarrow \left( { - x + 4} \right) \in B\left( 2 \right)\\
\to \left[ \begin{array}{l}
- x + 4 = 2\\
- x + 4 = 4\\
- x + 4 = 6\\
- x + 4 = 8\\
- x + 4 = 10\\
...
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 0\left( l \right)\\
x = - 2\\
x = - 4\\
x = - 6\\
...
\end{array} \right.
\end{array}\)
