Đáp án:
\(\begin{array}{l}
{A_{\max }} = 3 \Leftrightarrow x = 1\\
{B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow y = - \dfrac{3}{2}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 2x - {x^2} + 2\\
= 3 + \left( { - {x^2} + 2x - 1} \right)\\
= 3 - \left( {{x^2} - 2x + 1} \right)\\
= 3 - \left( {{x^2} - 2.x.1 + {1^2}} \right)\\
= 3 - {\left( {x - 1} \right)^2}\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\,\forall x\\
\Rightarrow A = 3 - {\left( {x - 1} \right)^2} \le 3,\,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 3 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1\\
\Rightarrow {A_{\max }} = 3 \Leftrightarrow x = 1\\
B = - {y^2} - 3y + \dfrac{1}{3}\\
= \dfrac{{31}}{{12}} + \left( { - {y^2} - 3y - \dfrac{9}{4}} \right)\\
= \dfrac{{31}}{{12}} - \left( {{y^2} + 3y + \dfrac{9}{4}} \right)\\
= \dfrac{{31}}{{12}} - \left[ {{y^2} + 2.y.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right]\\
= \dfrac{{31}}{{12}} - {\left( {y + \dfrac{3}{2}} \right)^2}\\
{\left( {y + \dfrac{3}{2}} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow B = \dfrac{{31}}{{12}} - {\left( {y + \dfrac{3}{2}} \right)^2} \le \dfrac{{31}}{{12}},\,\,\,\forall y\\
\Rightarrow {B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow {\left( {y + \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow y + \dfrac{3}{2} = 0 \Leftrightarrow y = - \dfrac{3}{2}\\
\Rightarrow {B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow y = - \dfrac{3}{2}
\end{array}\)
