Đáp án:
$\begin{array}{l}
a)A = {x^2} - 2x + 5\\
= {x^2} - 2x + 1 + 4\\
= {\left( {x - 1} \right)^2} + 4 \ge 4\\
\Leftrightarrow A \ge 4\\
\Leftrightarrow GTNN:A = 4\,khi:x = 1\\
b)B = {x^2} - x + 1\\
= {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\Leftrightarrow B \ge \dfrac{3}{4}\\
\Leftrightarrow GTNN:B = \dfrac{3}{4}\,khi:x = \dfrac{1}{2}\\
c)C = \left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
= \left( {x - 1} \right)\left( {x + 6} \right)\left( {x + 2} \right)\left( {x + 3} \right)\\
= \left( {{x^2} + 5x - 6} \right).\left( {{x^2} + 5x + 6} \right)\\
= {\left( {{x^2} + 5x} \right)^2} - 36 \ge - 36\\
\Leftrightarrow GTNN:C = - 36\\
khi:{x^2} + 5x = 0 \Leftrightarrow x = 0;x = - 5\\
d)D = {x^2} + 5{y^2} - 2xy + 4y + 3\\
= {x^2} - 2xy + {y^2} + 4{y^2} + 4y + 1 + 2\\
= {\left( {x - y} \right)^2} + {\left( {2y + 1} \right)^2} + 2 \ge 2\\
\Leftrightarrow GTNN:D = 2\,khi:\left\{ \begin{array}{l}
x = y\\
y = - \dfrac{1}{2}
\end{array} \right.
\end{array}$
