Đáp án:
$MaxB = \dfrac{{57}}{8} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{3}{4};\dfrac{{ - 1}}{4}} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{\left( {x - 1} \right)^3} + x - 1 = {y^3} + y\\
\Leftrightarrow {\left( {x - 1} \right)^3} - {y^3} + \left( {x - 1} \right) - y = 0\\
\Leftrightarrow \left( {x - 1 - y} \right)\left( {{{\left( {x - 1} \right)}^2} + \left( {x - 1} \right)y + {y^2}} \right) + \left( {x - 1 - y} \right) = 0\\
\Leftrightarrow \left( {x - 1 - y} \right)\left( {{{\left( {x - 1} \right)}^2} + \left( {x - 1} \right)y + {y^2} + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 - y = 0\\
{\left( {x - 1} \right)^2} + \left( {x - 1} \right)y + {y^2} + 1 = 0\left( {mt} \right)
\end{array} \right.\\
\Leftrightarrow x - 1 - y = 0\\
\Leftrightarrow x = y + 1
\end{array}$
Khi đó:
$\begin{array}{l}
B = - {x^2} - {y^2} + 3x - 2y + 5\\
= - {\left( {y + 1} \right)^2} - {y^2} + 3\left( {y + 1} \right) - 2y + 5\\
= - 2{y^2} - y + 7\\
= - 2\left( {{y^2} + 2.y.\dfrac{1}{4} + \dfrac{1}{{16}}} \right) + \dfrac{{57}}{8}\\
= - 2{\left( {y + \dfrac{1}{4}} \right)^2} + \dfrac{{57}}{8}\\
\le \dfrac{{57}}{8},\forall y \in R
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {y + \dfrac{1}{4}} \right)^2} = 0 \Leftrightarrow y = \dfrac{{ - 1}}{4} \Rightarrow x = y + 1 = \dfrac{3}{4}$
Vậy $MaxB = \dfrac{{57}}{8} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{3}{4};\dfrac{{ - 1}}{4}} \right)$
