Đáp án:
\(\begin{array}{l}
a)\,\,\min A = \frac{{11}}{4} \Leftrightarrow x = \frac{3}{2}\\
b)\,\,\min B = 1977 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,A = {x^2} - 3x + 5\\
A = {x^2} - 2x.\frac{3}{2} + \frac{9}{4} - \frac{9}{4} + 5\\
A = {\left( {x - \frac{3}{2}} \right)^2} + \frac{{11}}{4}\\
\Rightarrow A \ge \frac{{11}}{4}\\
\Rightarrow \min A = \frac{{11}}{4} \Leftrightarrow x = \frac{3}{2}\\
b)\,\,B = \left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 2} \right)\left( {x + 6} \right) + 2013\\
\,\,\,\,\,\,B = \left( {x - 1} \right)\left( {x + 6} \right)\left( {x + 3} \right)\left( {x + 2} \right) + 2013\\
\,\,\,\,\,\,B = \left( {{x^2} + 5x - 6} \right)\left( {{x^2} + 5x + 6} \right) + 2013\\
Dat\,\,t = {x^2} + 5x\\
\Rightarrow B = \left( {t - 6} \right)\left( {t + 6} \right) + 2013\\
\,\,\,\,\,\,B = {t^2} - 36 + 2013\\
\,\,\,\,\,\,B = {t^2} + 1977\\
\Rightarrow B \ge 1977\\
\Rightarrow \min B = 1977 \Leftrightarrow t = 0\\
\Rightarrow {x^2} + 5x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.
\end{array}\)
