Đáp án:
x=0
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 1\\
A = \dfrac{{3\left( {x + 1} \right)}}{{{x^2}\left( {x - 1} \right) + \left( {x - 1} \right)}}\\
= \dfrac{{3\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{{3\left( {x - 1 + 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{{3\left( {x - 1} \right) + 6}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{3}{{{x^2} + 1}} + \dfrac{6}{{{x^3} - {x^2} + x - 1}}\\
Để:A \in Z\\
\to \left\{ \begin{array}{l}
\dfrac{3}{{{x^2} + 1}} \in Z\\
\dfrac{6}{{{x^3} - {x^2} + x - 1}} \in Z
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} + 1 \in U\left( 3 \right)\\
{x^3} - {x^2} + x - 1 \in U\left( 6 \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{x^2} + 1 = 3\\
{x^2} + 1 = 1
\end{array} \right.\\
\left[ \begin{array}{l}
{x^3} - {x^2} + x - 1 = 6\\
{x^3} - {x^2} + x - 1 = - 6\\
{x^3} - {x^2} + x - 1 = 3\\
{x^3} - {x^2} + x - 1 = - 3\\
{x^3} - {x^2} + x - 1 = 2\\
{x^3} - {x^2} + x - 1 = - 2\\
{x^3} - {x^2} + x - 1 = 1\\
{x^3} - {x^2} + x - 1 = - 1
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{x^2} = 2\\
{x^2} = 0
\end{array} \right.\\
\left[ \begin{array}{l}
x = 2,104872786\\
x = - 1,278163073\\
x = 1,742959202\\
x = - 0,8105357138\\
x = - 0,5436890127\\
x = 1,353209964\\
x = 0
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = \pm \sqrt 2 \\
x = 0
\end{array} \right.\\
\left[ \begin{array}{l}
x = 2,104872786\\
x = - 1,278163073\\
x = 1,742959202\\
x = - 0,8105357138\\
x = - 0,5436890127\\
x = 1,353209964\\
x = 0
\end{array} \right.
\end{array} \right.\\
KL:x = 0
\end{array}\)
