Đáp án:
a) $MinA = - 1 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 5 - \sqrt 5 }}{2}
\end{array} \right.$
b) $MinB = - 3 \Leftrightarrow \left( {x;y} \right) = \left( {1;1} \right)$
Giải thích các bước giải:
$\begin{array}{l}
a)A = \left( {{x^2} + 3x + 2} \right)\left( {{x^2} + 7x + 12} \right)\\
= \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\\
= \left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)\left( {x + 3} \right)\\
= \left( {{x^2} + 5x + 4} \right)\left( {{x^2} + 5x + 6} \right)\\
= {\left( {{x^2} + 5x} \right)^2} + 10\left( {{x^2} + 5x} \right) + 24\\
= {\left( {{x^2} + 5x} \right)^2} + 10\left( {{x^2} + 5x} \right) + 25 - 1\\
= {\left( {{x^2} + 5x + 5} \right)^2} - 1
\end{array}$
Mà ta có:
$\begin{array}{l}
{\left( {{x^2} + 5x + 5} \right)^2} \ge 0,\forall x\\
\Rightarrow A = {\left( {{x^2} + 5x + 5} \right)^2} - 1 \ge - 1,\forall x
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow {\left( {{x^2} + 5x + 5} \right)^2} = 0\\
\Leftrightarrow {x^2} + 5x + 5 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 5 - \sqrt 5 }}{2}
\end{array} \right.
\end{array}$
Vậy $MinA = - 1 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 5 - \sqrt 5 }}{2}
\end{array} \right.$
b) Ta có:
$\begin{array}{l}
B = {x^2} + xy + {y^2} - 3x - 3y\\
= {x^2} + x\left( {y - 3} \right) + {y^2} - 3y\\
= {x^2} + 2.x.\dfrac{{y - 3}}{2} + \dfrac{1}{4}{\left( {y - 3} \right)^2} + \dfrac{3}{4}{y^2} - \dfrac{3}{2}y - \dfrac{9}{4}\\
= {\left( {x + \dfrac{{y - 3}}{2}} \right)^2} + \dfrac{3}{4}\left( {{y^2} - 2y + 1} \right) - 3\\
= {\left( {x + \dfrac{{y - 3}}{2}} \right)^2} + \dfrac{3}{4}{\left( {y - 1} \right)^2} - 3
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + \dfrac{{y - 3}}{2}} \right)^2} \ge 0\\
{\left( {y - 1} \right)^2} \ge 0
\end{array} \right.,\forall x,y\\
\Rightarrow {\left( {x + \dfrac{{y - 3}}{2}} \right)^2} + \dfrac{3}{4}{\left( {y - 1} \right)^2} - 3 \ge - 3,\forall x,y\\
\Rightarrow B \ge - 3,\forall x,y
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + \dfrac{{y - 3}}{2}} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{{y - 3}}{2} = 0\\
y - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 1\\
x = 1
\end{array} \right.
\end{array}$
Vậy $MinB = - 3 \Leftrightarrow \left( {x;y} \right) = \left( {1;1} \right)$
