Đáp án:
c. \(\left[ \begin{array}{l}
x = 1\\
x = - 5\\
x = - 1\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x + 2 \ne 0 \to x \ne - 2\\
b.C = \dfrac{{3{x^2} + 6x - 10x - 20 + 3}}{{x + 2}}\\
= \dfrac{{3x\left( {x + 2} \right) - 10\left( {x + 2} \right) + 3}}{{x + 2}}\\
= 3x - 10 + \dfrac{3}{{x + 2}}\\
c.Để:C \in Z\\
\to \dfrac{3}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 3\\
x + 2 = - 3\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = - 5\\
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
