Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 4{x^2} - x + 5\\
= \left( {4{x^2} - x + \dfrac{1}{{16}}} \right) + \dfrac{{79}}{{16}}\\
= \left[ {{{\left( {2x} \right)}^2} - 2.2x.\dfrac{1}{4} + {{\left( {\dfrac{1}{4}} \right)}^2}} \right] + \dfrac{{79}}{{16}}\\
= {\left( {2x - \dfrac{1}{4}} \right)^2} + \dfrac{{79}}{{16}} \ge \dfrac{{79}}{{16}},\,\,\,\forall x\\
\Rightarrow {A_{\min }} = \dfrac{{79}}{{16}} \Leftrightarrow {\left( {2x - \dfrac{1}{4}} \right)^2} = 0 \Leftrightarrow 2x - \dfrac{1}{4} = 0 \Leftrightarrow x = \dfrac{1}{8}\\
B = 3x - {x^2} + 9\\
= \left( {3x - {x^2} - \dfrac{9}{4}} \right) + \dfrac{{45}}{4}\\
= - \left( {{x^2} - 3x + \dfrac{9}{4}} \right) + \dfrac{{45}}{4}\\
= \dfrac{{45}}{4} - \left( {{x^2} - 2.x.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right)\\
= \dfrac{{45}}{4} - {\left( {x - \dfrac{3}{2}} \right)^2} \le \dfrac{{45}}{4},\,\,\,\forall x\\
\Rightarrow {B_{\max }} = \dfrac{{45}}{4} \Leftrightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{2}\\
D = {x^2} + 3{y^2} + 6y - 4x - 1\\
= \left( {{x^2} - 4x + 4} \right) + \left( {3{y^2} + 6y + 3} \right) - 8\\
= {\left( {x - 2} \right)^2} + 3.\left( {{y^2} + 2y + 1} \right) - 8\\
= {\left( {x - 2} \right)^2} + 3.{\left( {y + 1} \right)^2} - 8 \ge - 8,\,\,\,\forall x,y\\
\Rightarrow {D_{\min }} = - 8 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = - 1
\end{array} \right.
\end{array}\)
Em xem lại đề câu c nhé!
