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