Đáp án:
min A=2
min B$=- \dfrac{9}{2}$
min C=5
min D=-1
min E=2013
Giải thích các bước giải:
\(\begin{array}{l}
A = {x^2} - 4x + 6 = {\left( {x - 2} \right)^2} + 2 \ge 2\\
{A_{\min }} = 2 \Leftrightarrow x = 2\\
B = 2\left( {{x^2} + 3x} \right) = 2{\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{9}{2} \ge - \dfrac{9}{2}\\
{B_{\min }} = - \dfrac{9}{2} \Leftrightarrow x + \dfrac{3}{2} = 0 \Leftrightarrow x = - \dfrac{3}{2}\\
C = 4{x^2} - 4x + 1 + {x^2} + 4x + 4\\
= 5{x^2} + 5 \ge 5 \Rightarrow {C_{\min }} = 5 \Leftrightarrow x = 0\\
D = \left( {{x^2} + 3x} \right)\left( {{x^2} + 3x + 2} \right)\\
Dat\,{x^2} + 3x + 1 = t \Rightarrow D = \left( {t - 1} \right)\left( {t + 1} \right) = {t^2} - 1 \ge - 1\\
{D_{\min }} = - 1 \Leftrightarrow t = 0 \Leftrightarrow {x^2} + 3x + 1 = 0 \Leftrightarrow x = \dfrac{{ - 3 \pm \sqrt 5 }}{2}\\
E = \left( {{x^2} + 4{y^2} + 1 + 2x - 4xy - 4y} \right) + {x^2} - 6x + 9 + 2013\\
= \left( {x - 2y + 1} \right) + {\left( {x - 3} \right)^2} + 2013 \ge 2013\\
{E_{\min }} = 2013 \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
x - 2y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.
\end{array}\)
