Đáp án:
\(Min = \dfrac{3}{4}\)
Giải thích các bước giải:
b) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to {m^2} - \left( {m - 1} \right)\left( {m + 1} \right) \ge 0\\
\to {m^2} - {m^2} + 1 \ge 0\\
\to 1 \ge 0\left( {ld} \right)\\
B = {x_1}^2 + {x_2}^2 + {x_1}{x_2}\\
= {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - {x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - {x_1}{x_2}\\
= {\left( {\dfrac{{ - 2m}}{{m - 1}}} \right)^2} - \dfrac{{m + 1}}{{m - 1}}\\
= \dfrac{{4{m^2} - \left( {m + 1} \right)\left( {m + 1} \right)}}{{{{\left( {m - 1} \right)}^2}}}\\
= \dfrac{{4{m^2} - {m^2} + 1}}{{{{\left( {m - 1} \right)}^2}}}\\
= \dfrac{{3{m^2} + 1}}{{{{\left( {m - 1} \right)}^2}}}\\
= \dfrac{{3\left( {{m^2} - 2m + 1} \right) + 6m - 2}}{{{{\left( {m - 1} \right)}^2}}}\\
= \dfrac{{3{{\left( {m - 1} \right)}^2} + 6\left( {m - 1} \right) + 4}}{{{{\left( {m - 1} \right)}^2}}}\\
= 3 + \dfrac{6}{{m - 1}} + \dfrac{4}{{{{\left( {m - 1} \right)}^2}}}\\
Đặt:\dfrac{1}{{m - 1}} = t\\
Pt \to 3 + 6t + 4{t^2} = 4{t^2} + 2.2t.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}\\
= {\left( {2t + \dfrac{3}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {2t + \dfrac{3}{2}} \right)^2} \ge 0\forall t\\
\to {\left( {2t + \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to Min = \dfrac{3}{4}\\
\Leftrightarrow 2t + \dfrac{3}{2} = 0\\
\to t = - \dfrac{3}{4}\\
\to \dfrac{1}{{m - 1}} = - \dfrac{3}{4}\\
\to - 4 = 3m - 3\\
\to m = - \dfrac{1}{3}
\end{array}\)