Đáp án:
b) m=3
Giải thích các bước giải:
\(\begin{array}{l}
a)Xét:\Delta ' = {m^2} - 6m + 9 + 4m - 8\\
= {m^2} - 2m + 1\\
= {\left( {m - 1} \right)^2} \ge 0\forall m\\
\to dpcm\\
b)DK:\Delta ' > 0\\
\to m - 1 \ne 0 \to m \ne 1\\
Có:A = \left( {\dfrac{{{x_1}}}{{{x_2}}} + 1} \right)\left( {\dfrac{{{x_2}}}{{{x_1}}} + 1} \right)\\
= \dfrac{{{x_1}}}{{{x_2}}}.\dfrac{{{x_2}}}{{{x_1}}} + \dfrac{{{x_1}}}{{{x_2}}} + \dfrac{{{x_2}}}{{{x_1}}} + 1\\
= \dfrac{{{x_1}}}{{{x_2}}} + \dfrac{{{x_2}}}{{{x_1}}} + 2\\
= \dfrac{{{x_1}^2 + {x_2}^2}}{{{x_1}{x_2}}} + 2\\
= \dfrac{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}}}{{{x_1}{x_2}}} + 2\\
= \dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}{{{x_1}{x_2}}} + 2\\
= \dfrac{{{{\left( {2m - 6} \right)}^2} - 2\left( { - 4m + 8} \right)}}{{ - 4m + 8}} + 2\\
= \dfrac{{4{m^2} - 24m + 36 + 8m - 16 - 8m + 16}}{{8 - 4m}}\\
= \dfrac{{4{m^2} - 24m + 36}}{{4\left( {2 - m} \right)}}\\
= \dfrac{{4\left( {4 - 4m + {m^2}} \right) - 8m + 20}}{{4\left( {2 - m} \right)}}\\
= \dfrac{{4{{\left( {2 - m} \right)}^2} + 8\left( {2 - m} \right) + 4}}{{4\left( {2 - m} \right)}}\\
= \left( {2 - m} \right) + 2 + \dfrac{1}{{2 - m}}\\
A \in Z \Leftrightarrow \dfrac{1}{{2 - m}} \in Z\\
\Leftrightarrow 2 - m \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
2 - m = 1\\
2 - m = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 1\left( l \right)\\
m = 3
\end{array} \right.
\end{array}\)
