Đáp án:
$\begin{array}{l}
A = {x^2} - 12x + 7\\
= {x^2} - 2.6.x + 36 - 29\\
= {\left( {x - 6} \right)^2} - 29 \ge - 29\\
\Leftrightarrow A \ge - 29\\
\Leftrightarrow GTNN:A = - 29\\
\text{Dấu = xảy ra khi}:x = 6\\
B = {x^2} + x + 2\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\\
\Leftrightarrow B \ge \dfrac{7}{4}\\
\Leftrightarrow GTNN:B = \dfrac{7}{4}\\
\text{Dấu = xảy ra khi}:x = - \dfrac{1}{2}\\
C = \dfrac{{1 - 4x}}{{{x^2}}}\\
\Leftrightarrow C.{x^2} = 1 - 4x\\
\Leftrightarrow C.{x^2} + 4x - 1 = 0\\
\Leftrightarrow \Delta ' \ge 0\\
\Leftrightarrow 4 + C \ge 0\\
\Leftrightarrow C \ge - 4\\
\Leftrightarrow GTNN:C = - 4\\
\text{Dấu = xảy ra khi}:\dfrac{{1 - 4x}}{{{x^2}}} = - 4\\
\Leftrightarrow - 4{x^2} = - 4x + 1\\
\Leftrightarrow 4{x^2} - 4x + 1 = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 0\\
\Leftrightarrow x = \dfrac{1}{2}\\
D = x + y\left( {x,y > 0;x.y = 25} \right)\\
= x + \dfrac{{25}}{x}\left( {do:x.y = 25} \right)\\
Theo\,Co - si:x + \dfrac{{25}}{x} \ge 2\sqrt {x.\dfrac{{25}}{x}} = 10\\
\Leftrightarrow D \ge 10\\
\Leftrightarrow GTNN:D = 10\\
\text{Dấu = xảy ra khi}:x = \dfrac{{25}}{x}\\
\Leftrightarrow {x^2} = 25\\
\Leftrightarrow x = y = 5\\
E = {x^3} + {y^3};x + y = 2\\
= \left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right)\\
= 2.\left( {{x^2} - xy + {y^2}} \right)\left( {do:x + y = 2} \right)\\
= 2.\left( {{x^2} - x.\left( {2 - x} \right) + {{\left( {2 - x} \right)}^2}} \right)\\
= 2.\left( {{x^2} - 2x + {x^2} + {x^2} - 4x + 4} \right)\\
= 2.\left( {3{x^2} - 6x + 4} \right)\\
= 2.3.\left( {{x^2} - 2x + \dfrac{4}{3}} \right)\\
= 6.\left( {{x^2} - 2x + 1 + \dfrac{1}{3}} \right)\\
= 6.{\left( {x - 1} \right)^2} + 6.\dfrac{1}{3} \ge 2\\
\Leftrightarrow GTNN:E = 2\\
\text{Dấu = xảy ra khi}:x = 1 \Leftrightarrow y = 1
\end{array}$
