Đáp án:
b. \(MinP = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = 3\\
Pt \to {x^2} - 2x - 1 = 0\\
\to {\left( {x - 1} \right)^2} = 2\\
\to \left[ \begin{array}{l}
x - 1 = \sqrt 2 \\
x - 1 = - \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 2 \\
x = 1 - \sqrt 2
\end{array} \right.\\
b.P = {\left( {{x_1}{x_2}} \right)^2} + 3\left( {{x_1}^2 + {x_2}^2} \right) - 4\\
= {\left( {{x_1}{x_2}} \right)^2} + 3\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) - 4\\
= {\left( {{x_1}{x_2}} \right)^2} + 3{\left( {{x_1} + {x_2}} \right)^2} - 6{x_1}{x_2} - 4\\
= {\left( {2 - m} \right)^2} + {3.2^2} - 6\left( {2 - m} \right) - 4\\
= 4 - 4m + {m^2} + 8 - 12 + 6m\\
= {m^2} + 2m\\
= {m^2} + 2m + 1 - 1\\
= {\left( {m + 1} \right)^2} - 1\\
Do:{\left( {m + 1} \right)^2} \ge 0\forall m \in R\\
\to {\left( {m + 1} \right)^2} - 1 \ge - 1\\
\to P \ge - 1\\
\to MinP = - 1\\
\Leftrightarrow m + 1 = 0\\
\Leftrightarrow m = - 1
\end{array}\)
