Đáp án:
b) \(MinB = \dfrac{{11}}{3}\)
Giải thích các bước giải:
a) Để phương trình có 2 nghiệm trái dấu
\(\begin{array}{l}
\Leftrightarrow - {a^2} + a - 2 < 0\\
\to - \left( {{a^2} - 2a.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{2}} \right) < 0\\
\to - {\left( {a - \dfrac{1}{2}} \right)^2} - \dfrac{7}{2} < 0\left( {ld} \right)\forall a \in R\\
b)Xét:\Delta \ge 0\\
\to {a^2} - 2a + 1 + 4{a^2} - 4a + 8 \ge 0\\
\to 5{a^2} - 6a + 9 \ge 0\left( {ld} \right)\forall a \in R\\
B = {x_1}^2 + {x_2}^2\\
= \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right)\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( {a - 1} \right)^2} - 2\left( { - {a^2} + a - 2} \right)\\
= {a^2} - 2a + 1 + 2{a^2} - 2a + 4\\
= 3{a^2} - 4a + 5\\
= 3{a^2} - 2.a\sqrt 3 .\dfrac{2}{{\sqrt 3 }} + \dfrac{4}{3} + \dfrac{{11}}{3}\\
= {\left( {a\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{{11}}{3}\\
Do:{\left( {a\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} \ge 0\forall a\\
\to {\left( {a\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{{11}}{3} \ge \dfrac{{11}}{3}\\
\to MinB = \dfrac{{11}}{3}\\
\Leftrightarrow a\sqrt 3 - \dfrac{2}{{\sqrt 3 }} = 0\\
\to a = \dfrac{2}{3}
\end{array}\)
