\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)

Ta có : \(\left(2x-5\right)^{2012}\ge0\forall x\)

            \(\left(3y+4\right)^{2014}\ge0\forall y\)

\(\rightarrow\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\ge0\forall x,y\)

Theo bài : \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0\)

\(\rightarrow\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}=0\)

\(\rightarrow\left(2x-5\right)^{2012}=0,\left(3y+4\right)^{2014}=0\)

\(\rightarrow2x-5=0,3y+4=0\)

\(\rightarrow x=\frac{5}{2};y=\frac{-4}{3}\)

Tự tìm M nhé bạn

1, M + (5x²-2xy)= 6x²+9xy-y²

    M                    =(6x²+9xy-y²)- (5x²-2xy)

    M                    = 6x²+9xy-y²-5x²+2xy

    M                    = (6x²-5x²)+(9xy+2xy)-y²

    M                    = x²+11xy-y²

* M + ( 5x² - 2xy ) = 6x² + 9xy - y²

<=> M = ( 6x² + 9xy - y² ) - ( 5x² - 2xy )

<=> M = 6x² + 9xy - y² - 5x² + 2xy

<=> M = x² + 11xy - y²

* \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2012}\ge0\forall x\\\left(3y+4\right)^{2014}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\ge0\)

Dấu = xảy ra <=> \(\hept{\begin{cases}\left(2x-5\right)^{2012}=0\\\left(3y+4\right)^{2014}=0\end{cases}}\)

                     <=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\)

                     <=> \(\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy M = -1159/36 khi x = 5/2 ; y = -4/3

Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\left(\forall x\right)\\\left(3y+4\right)^{2020}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\left(\forall x,y\right)\)

Mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\left(\forall x,y\right)\)

\(\Rightarrow\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Khi đó thay vào ta được: 

\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)

\(\Rightarrow M=-\frac{1159}{36}\)

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\)   \(\forall x,y\)

mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)  (đề bài ) \(\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

\(\Rightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)

Rút gọn biểu thức

\(m+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

=> \(m=x^2+11xy-y^2\)

Thay x,y, vừa tìm được vào biểu thức đã được rút gọn ta tính được m 

Đây là bài hướng dẫn, có gì thắc mắc hãy hỏi lại!!

\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)

\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)

Ta có: \(\hept{\begin{cases}\left(2021x-1\right)^{2020}\ge0\\\left(3y+4\right)^{2022}\ge0\end{cases}}\left(\forall x,y\right)\)

\(\Rightarrow\left(2021x-1\right)^{2020}+\left(3y+4\right)^{2022}\ge0\left(\forall x,y\right)\)

Mà theo đề bài ta có: \(\left(2021x-1\right)^{2020}+\left(3y+4\right)^{2022}\le0\)

Nên từ đó suy ra: \(\hept{\begin{cases}\left(2021x-1\right)^{2020}=0\\\left(3y+4\right)^{2022}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2021x-1=0\\3y+4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2021}\\y=-\frac{4}{3}\end{cases}}\)

Khi đó \(M=2021\cdot\frac{1}{2021}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(=-\frac{4}{3}-\frac{16}{9}=-\frac{28}{9}\)