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

M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2
(2x-5)^2020+(3y+4)^2022<=0

=>x=5/2 và y=-4/3

M=25/4+11*5/2*(-4/3)-16/9=-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

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

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

\(M=6x^2+9xy-y^2-5x^2+2xy\)

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

\(M=x^2+11xy-y^2\)

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

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

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

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

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

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 giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

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

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

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=x^2+11xy-y^2\)

\(N=3xy-4y^2-x^2+7xy-8y^2\)

\(N=-x^2+10xy-12y^2\)

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

\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)

b.     \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)

\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)