Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
A = 4{a^2} + 4a + 2 = \left( {4{a^2} + 4a + 1} \right) + 1 = {\left( {2a + 1} \right)^2} + 1 \ge 1,\,\,\forall a\\
\Rightarrow {A_{\min }} = 1 \Leftrightarrow {\left( {2a + 1} \right)^2} = 0 \Leftrightarrow a = - \frac{1}{2}\\
*)\\
B = {x^2} - 4xy + 5{y^2} + 10x - 22y + 28\\
= \left( {{x^2} - 4xy + 4{y^2}} \right) + \left( {10x - 20y} \right) + \left( {{y^2} - 2y + 1} \right) + 27\\
= {\left( {x - 2y} \right)^2} + 10\left( {x - 2y} \right) + {\left( {y - 1} \right)^2} + 27\\
= \left[ {{{\left( {x - 2y} \right)}^2} + 2.\left( {x - 2y} \right).5 + 25} \right] + {\left( {y - 1} \right)^2} + 2\\
= {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2,\,\,\,\forall x,y\\
\Rightarrow {B_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
x - 2y + 5 = 0\\
y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 3\\
y = 1
\end{array} \right.\\
*)\\
C = {x^2} + 3x + 7\\
= \left( {{x^2} + 2.x.\frac{3}{2} + \frac{9}{4}} \right) + \frac{{19}}{4}\\
= {\left( {x + \frac{3}{2}} \right)^2} + \frac{{19}}{4} \ge \frac{{19}}{4},\,\,\,\forall x\\
\Rightarrow {C_{\min }} = \frac{{19}}{4} \Leftrightarrow x + \frac{3}{2} = 0 \Leftrightarrow x = - \frac{3}{2}
\end{array}\)
