Đáp án:
3) \(\left[ \begin{array}{l}
x = 4\\
x = - 2\\
x = 2\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \left[ {\dfrac{{{x^2} + 1 + {x^2}}}{{{x^2} + 1}}} \right]:\left[ {\dfrac{{{x^2} + 1 - 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}} \right]\\
= \dfrac{{2{x^2} + 1}}{{{x^2} + 1}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{2{x^2} + 1}}{{x - 1}}\\
2)Thay:x = - \dfrac{1}{2}\\
\to A = \dfrac{{2.{{\left( { - \dfrac{1}{2}} \right)}^2} + 1}}{{ - \dfrac{1}{2} - 1}} = - 1\\
3)A = \dfrac{{2{x^2} + 1}}{{x - 1}} = \dfrac{{2\left( {{x^2} - 2x + 1} \right) + 4x - 1}}{{x - 1}}\\
= \dfrac{{2{{\left( {x - 1} \right)}^2} + 4\left( {x - 1} \right) + 3}}{{x - 1}}\\
= 2\left( {x - 1} \right) + 4 + \dfrac{3}{{x - 1}}\\
Để:A \in Z\\
\Leftrightarrow \dfrac{3}{{x - 1}} \in Z\\
\Leftrightarrow x - 1 \in U\left( 3 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - 2\\
x = 2\\
x = 0
\end{array} \right.
\end{array}\)
