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\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\) \(\forall x,y\)
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
\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
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
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:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)
a) Ta có: \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
Vậy: \(M=x^2+11xy-y^2\)
b) Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)
Vậy: \(N=-x^2+10xy-12y^2\)
a, (6x2+9xy-y2) - ( 5x2-2xy)=M
=> M= (6x2+9xy-y2) - ( 5x2-2xy)
=> M= 6x2+9xy-y2 - 5x2+2xy
=> M=(6x2- 5x2)+(9xy+2xy)-y2
=>M= 1x2 + 11xy - y2
Vậy M= 1x2 + 11xy - y2
b, N= (3xy-4y2) - (x2-7xy+8y2)
=> N= 3xy-4y2 - x2+7xy-8y2
=> N= (3xy+7xy)-(4y2+8y2)-x2
=> N= 10xy - 12y2 -x2
Vậy N= 10xy - 12y2 -x2
a: Ta có: \(M+5x^2-2xy=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
b: Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)
