\(A=-2x^2+4xy-2y^2+4\left(x-y\right)-2-8y^2+8y+2019\\ A=\left[-2\left(x-y\right)^2+4\left(x-y\right)-2\right]-8\left(y^2-y+\dfrac{1}{4}\right)+2020\\ A=-2\left(x-y-1\right)^2-8\left(y-\dfrac{1}{2}\right)^2+2020\le2020\\ A_{max}=2020\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+\dfrac{1}{2}=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=-2x^2-10y^2+4xy+4x+4y+2016\)

\(=-2.\left(x^2+5y^2-4xy-4x-4y\right)+2016\)

\(=-2.\left(x^2+4y^2+4-4xy-4x+8y+y^2-12y+36\right)+2.36+2016\)

\(=-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]+2088\)

Ta có: \(\left(x-2y-2\right)^2+\left(y-6\right)^2\ge0\)

\(\Rightarrow-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]\le0\)

\(\Rightarrow-2.[\left(x-2y-2\right)^2+\left(y-6\right)^2]+2088\le2088\)

\(\Rightarrow A\le2088\)

Vậy giá trị lớn nhất của \(A=2088\) khi: \(\hept{\begin{cases}x-2y-2=0\\y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=2y+2\\y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=14\\y=6\end{cases}}\)

Ta có:

A=-2x^2-10y^2+4xy +4x+4y+2013

=-(2x^2+10^2-4xy-4x-4y-2013)

=-[(2x^2+2y^2-4xy)-(4x-4y)+2-2015+8y^2-8y]

=-[2(x-y)^2-4(x-y)+2+(8y^2-8y+2)-2017]

=-[2(x-y-1)^2+8(y-1/4)^2]+2017

vì 2(x-y-1)^2\(\ge\)0với mọi x,y

8(y-1/4)^2\(\ge\)0với mọi y

=>-[2(x-y-1)^2+8(y-1/4)^2]\(\le\)0với mọi x,y

=>A=-[2(x-y-1)^2+8(y-1/4)^2]+2017\(\le\)2017với mọi x,y

dấu "=" xảy ra khi\(\Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{1}{4}=0\\x-y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\x-\dfrac{5}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Vậy GTLN của A là 2017 khi y=1/4;x=5/4

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn