Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {x^2} - 5x + 12\\
= \left( {{x^2} - 5x + \dfrac{{25}}{4}} \right) + \dfrac{{23}}{4}\\
= \left( {{x^2} - 2.x.\dfrac{5}{2} + {{\left( {\dfrac{5}{2}} \right)}^2}} \right) + \dfrac{{23}}{4}\\
= {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{{23}}{4} \ge \dfrac{{23}}{4},\,\,\,\forall x\\
\Rightarrow {A_{\min }} = \dfrac{{23}}{4} \Leftrightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{5}{2}\\
B = {x^2} + x + 1\\
= \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
= \left( {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
\Rightarrow {B_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = - \dfrac{1}{2}\\
C = {x^2} - 7x + 10\\
= \left( {{x^2} - 7x + \dfrac{{49}}{4}} \right) - \dfrac{9}{4}\\
= \left( {{x^2} - 2.x.\dfrac{7}{2} + {{\left( {\dfrac{7}{2}} \right)}^2}} \right) - \dfrac{9}{4}\\
= {\left( {x - \dfrac{7}{2}} \right)^2} - \dfrac{9}{4} \ge - \dfrac{9}{4},\,\,\,\forall x\\
\Rightarrow {C_{\min }} = - \dfrac{9}{4} \Leftrightarrow {\left( {x - \dfrac{7}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{7}{2}\\
D = {x^2} + 10x + 40\\
= \left( {{x^2} + 10x + 25} \right) + 15\\
= \left( {{x^2} + 2.x.5 + {5^2}} \right) + 15\\
= {\left( {x + 5} \right)^2} + 15 \ge 15,\,\,\,\,\forall x\\
\Rightarrow {D_{\min }} = 15 \Leftrightarrow {\left( {x + 5} \right)^2} = 0 \Leftrightarrow x = - 5
\end{array}\)
