Đáp án:
\(P = - 6038\) hoặc \(P = \dfrac{{6037}}{4}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2{x^2} + 2{y^2} = 5xy\\
\Leftrightarrow 2{x^2} - 5xy + 2{y^2} = 0\\
\Leftrightarrow \left( {2{x^2} - 4xy} \right) - \left( {xy - 2{y^2}} \right) = 0\\
\Leftrightarrow 2x\left( {x - 2y} \right) - y.\left( {x - 2y} \right) = 0\\
\Leftrightarrow \left( {2x - y} \right).\left( {x - 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - y = 0\\
x - 2y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = y\\
x = 2y
\end{array} \right.\\
TH1:\,\,\,\,\,2x = y\\
P = \dfrac{{2012x + 2013y}}{{3x - 2y}} = \dfrac{{2012x + 2013.2x}}{{3x - 2.2x}}\\
= \dfrac{{2012x + 4026x}}{{3x - 4x}} = \dfrac{{6038x}}{{ - x}} = - 6038\\
TH2:\,\,\,\,x = 2y\\
P = \dfrac{{2012x + 2013y}}{{3x - 2y}} = \dfrac{{2012.2y + 2013y}}{{3.2y - 2y}}\\
= \dfrac{{6037y}}{{4y}} = \dfrac{{6037}}{4}
\end{array}\)
Vậy \(P = - 6038\) hoặc \(P = \dfrac{{6037}}{4}\)
