Đáp án:
c. \(\left[ \begin{array}{l}
x = 0\\
x = - 6\\
x = - 2\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.Q = \left[ {\dfrac{{2{x^2} + 1 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right]:\left( {\dfrac{{{x^2} + x + 1 - {x^2} + 2}}{{{x^2} + x + 1}}} \right)\\
= \dfrac{{{x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
= \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
= \dfrac{x}{{x + 3}}\\
b.Thay:x = 2\\
\to Q = \dfrac{2}{{2 + 3}} = \dfrac{2}{5}\\
c.Q = \dfrac{{x + 3 - 3}}{{x + 3}} = 1 - \dfrac{3}{{x + 3}}\\
Q \in Z\\
\Leftrightarrow \dfrac{3}{{x + 3}} \in Z\\
\Leftrightarrow x + 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 3\\
x + 3 = - 3\\
x + 3 = 1\\
x + 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 6\\
x = - 2\\
x = - 4
\end{array} \right.
\end{array}\)
