Tìm GTNN của: a) (x² + 3x 2)(x² + 7x + 12) b) x² + xy + y² – 3x – 3y

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 $MinB =  – 3 \Leftrightarrow \left( {x;y} \right) = \left( {1;1} \right)$