*Lời giải :
Ta có : `(2x - 5)^{2018} + (3y + 4)^{2020} ≤ 0`
Vì \(\left\{ \begin{array}{l}(2x-5)^{2018}≥0∀x\\(3y+4)^{2020}≥0∀y\end{array} \right.\)
`-> (2x - 5)^{2018} + (3y + 4)^{2020} ≥0∀x,y`
Dấu "`=`" xảy ra khi :
\(\left\{ \begin{array}{l}2x-5=0\\3y+4=0\end{array} \right.\)
`->` \(\left\{ \begin{array}{l}2x=5\\3y=-4\end{array} \right.\)
`->` \(\left\{ \begin{array}{l}x=\dfrac{5}{2}\\y=\dfrac{-4}{3}\end{array} \right.\) (*)
Ta có : `M + (5x^2 - 2xy) = 6x^2 + 9xy - y^2`
`-> M = 6x^2 + 9xy - y^2 - 5x^2 + 2xy`
`-> M = (6x^2 - 5x^2) + (9xy + 2xy) - y^2`
`-> M = x^2 + 11xy - y^2`
Thay (*) vào `M` ta được :
`M = (5/2)^2 + 11 . 5/2 . (-4/3) - (-4/3)^2`
`-> M = (-1159)/36`
