Đáp án:
$\begin{array}{l}
A = {x^2} - 7x + 5\\
= {x^2} - 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} - \dfrac{{29}}{4}\\
= {\left( {x - \dfrac{7}{2}} \right)^2} - \dfrac{{29}}{4} \ge - \dfrac{{29}}{4}\\
\Rightarrow A \ge - \dfrac{{29}}{4}\\
\Rightarrow \min A = \dfrac{{ - 29}}{4}\,khi:x = \dfrac{7}{2}\\
B = 2{x^2} - 4x + 6\\
= 2\left( {{x^2} - 2x + 1} \right) + 4\\
= 2{\left( {x - 1} \right)^2} + 4 \ge 4\\
\Rightarrow B \ge 4\\
\Rightarrow \min B = 4\,Khi:x = 1\\
C = {x^2} + 3x - 2\\
= {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{{17}}{4}\\
= {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{4} \ge - \dfrac{{17}}{4}\\
\Rightarrow C \ge - \dfrac{{17}}{4}\\
\Rightarrow \min C = \dfrac{{ - 17}}{4}\,khi:x = - \dfrac{3}{2}\\
D = x - {x^2}\\
= - \left( {{x^2} - x} \right)\\
= - \left( {{x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= - {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\Rightarrow \max D = \dfrac{1}{4}\,khi:x = \dfrac{1}{2}\\
E = 13 - x - 2{x^2}\\
= - \left( {2{x^2} + x} \right) + 13\\
= - 2\left( {{x^2} + \dfrac{1}{2}x} \right) + 13\\
= - 2.\left( {{x^2} + 2.x.\dfrac{1}{4} + \dfrac{1}{{16}}} \right) + 2.\dfrac{1}{{16}} + 13\\
= - 2{\left( {x + \dfrac{1}{4}} \right)^2} + \dfrac{{105}}{8} \le \dfrac{{105}}{8}\\
\Rightarrow \max E = \dfrac{{105}}{8}\,khi:x = - \dfrac{1}{4}
\end{array}$
