Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
D = {x^2} + x + 2 = \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{7}{4}\\
= \left( {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{7}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4},\,\,\,\forall x\\
\Rightarrow {D_{\min }} = \dfrac{7}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = - \dfrac{1}{2}\\
e,\\
E = {x^2} - 3x + 5 = \left( {{x^2} - 3x + \dfrac{9}{4}} \right) + \dfrac{{11}}{4}\\
= \left( {{x^2} - 2.x.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right) + \dfrac{{11}}{4}\\
= {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{11}}{4} \ge \dfrac{{11}}{4},\,\,\,\,\forall x\\
\Rightarrow {E_{\min }} = \dfrac{{11}}{4} \Leftrightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{2}
\end{array}\)
