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