Đáp án:
b. \(MaxB = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = {x^2} + 6x + 9 - 18x + 3{x^2} + 3\\
= 4{x^2} - 12x + 12\\
= 4\left( {{x^2} - 3x + 3} \right)\\
= 4\left( {{x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}} \right)\\
= 4{\left( {x - \dfrac{3}{2}} \right)^2} + 3\\
Do:4{\left( {x - \dfrac{3}{2}} \right)^2} \ge 0\forall x \in R\\
\to 4{\left( {x - \dfrac{3}{2}} \right)^2} + 3 > 0\\
\to A > 0\\
B = 12x - 1 - {x^2} - 6x - 9\\
= - \left( {{x^2} - 6x + 10} \right)\\
= - \left( {{x^2} - 2.3.x + 9 + 1} \right)\\
= - {\left( {x - 3} \right)^2} - 1\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 3} \right)^2} \le 0\\
\to - {\left( {x - 3} \right)^2} - 1 < 0\\
\to B < 0\\
\to dpcm\\
b.A = 4{\left( {x - \dfrac{3}{2}} \right)^2} + 3\\
Do:4{\left( {x - \dfrac{3}{2}} \right)^2} \ge 0\forall x \in R\\
\to 4{\left( {x - \dfrac{3}{2}} \right)^2} + 3 \ge 3\\
\to Min = 3\\
\Leftrightarrow x = \dfrac{3}{2}\\
B = - {\left( {x - 3} \right)^2} - 1\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 3} \right)^2} \le 0\\
\to - {\left( {x - 3} \right)^2} - 1 \le - 1\\
\to Max = - 1\\
\Leftrightarrow x = 3
\end{array}\)
