Đáp án:
c) \(\left[ \begin{array}{l}
m = 3\\
m = - 5\\
m = - 3\\
m = 0\\
m = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{m^2}x + 4m - 3 = {m^2} + x\\
\to \left( {{m^2} - 1} \right)x = {m^2} - 4m + 3\\
\to \left( {m - 1} \right)\left( {m + 1} \right)x = \left( {m - 1} \right)\left( {m - 3} \right)\\
a)Thay:m = 2\\
Pt \to \left( {{2^2} - 1} \right)x = {2^2} - 4.2 + 3\\
\to 3x = - 1\\
\to x = - \dfrac{1}{3}\\
b)DK:\left( {m - 1} \right)\left( {m + 1} \right) \ne 0\\
\to m \ne \pm 1\\
\to x = \dfrac{{m - 3}}{{m + 1}}\\
c)x = \dfrac{{m - 3}}{{m + 1}} = \dfrac{{m + 1 - 4}}{{m + 1}} = 1 - \dfrac{4}{{m + 1}}\\
Do:x \in Z\\
\to \dfrac{4}{{m + 1}} \in Z\\
\to m + 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
m + 1 = 4\\
m + 1 = - 4\\
m + 1 = 2\\
m + 1 = - 2\\
m + 1 = 1\\
m + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
m = 3\\
m = - 5\\
m = 1\left( l \right)\\
m = - 3\\
m = 0\\
m = - 2
\end{array} \right.
\end{array}\)
