Đáp án:
a) \(\dfrac{{x - 1}}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 1;0} \right\}\\
A = \left[ {\dfrac{{\left( {x + 2} \right)\left( {x + 1} \right) + 6x - 9x\left( {x + 1} \right)}}{{3x\left( {x + 1} \right)}}} \right].\dfrac{{x + 1}}{{2\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{{x^2} + 3x + 2 + 6x - 9{x^2} - 9x}}{{3x\left( {x + 1} \right)}}.\dfrac{{x + 1}}{{2\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{ - 8{x^2} + 2}}{{3x}}.\dfrac{1}{{2\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{ - 2\left( {4{x^2} - 1} \right)}}{{6x\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{2\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{6x\left( {2x - 1} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{2x + 1}}{{3x}} - \dfrac{{3x - {x^2} + 1}}{{3x}} = \dfrac{{{x^2} - x}}{{3x}} = \dfrac{{x - 1}}{3}\\
b)A \in Z\\
\Leftrightarrow x - 1 \vdots 3\\
\to x - 1 \in B\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = 9\\
x - 1 = 12\\
x - 1 = 18\\
...
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 10\\
x = 13\\
x = 19\\
...
\end{array} \right.
\end{array}\)
