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\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
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
*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)
\(M=x^2+11xy-y^2\)
* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
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 \(M=2021\cdot\frac{1}{2021}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(=-\frac{4}{3}-\frac{16}{9}=-\frac{28}{9}\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=x^2+11xy-y^2\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=-x^2+10xy-12y^2\)
a. \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
b. \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
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:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)