A= (4x²+8xy+4y²)+ (x²-2x+1)-1+(y²+2y+1)-1+2019= 4(x+y)² + (x-1)²+(y+1)²+2017 \(\ge\)2017

Dấu "=" xảy ra khi      \(\hept{\begin{cases}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-y\\x=1\\y=-1\end{cases}}\)

Vậy MinA= 2017 khi x=1; y=-1

A=5+ (-x²+2x) +(-4y²-4y)= -(x²-2x+1)+1-(4y²+4y+1)+1+5=-(x-1)²-(2y+1)² +7 \(\le\)7

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}\)

Vậy Max A bằng 7 khi x=1; y=-1/2

Ta có : \(A=5-8x-x^2\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\)

Vì \(-\left(x+4\right)^2\le0\forall x\in R\)

Nên : \(-\left(x+4\right)^2+21\ge21\forall x\in R\)

Vậy : \(A_{min}=21\) khi x = -4

\(F=-x^4+x^2-4y^2+2x-4y+2000.\)

\(=-x^4+2x^2-1-x^2+2x-1-4y^2-4y-1+2003\)

\(=-\left(x^2-1\right)^2-\left(x-1\right)^2-\left(2y+1\right)^2+2003\)

\(=-\left(x-1\right)^2\left(x+1\right)^2-\left(x-1\right)^2-\left(2y+1\right)^2+2003\)

\(\Rightarrow F_{min}=2003\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(2y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}}\)

Vậy \(F_{min}=2003\Leftrightarrow x=1;y=-\frac{1}{2}\)

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

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

\(A=\left(x^2-2x\cdot1+1\right)+\left(4y^2+4y\right)+1\)

\(A=\left(x-1\right)^2+4\left(y^2+y\right)+1\)

Do \(\left(x-1\right)^2>\) hoặc bằng 0 và \(4\left(y^2+y\right)\)> hoặc bằng 0

nên để A đạt GTNN thì \(\left\{{}\begin{matrix}x-1=0\\y^2+y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(A=5-x^2+2x-4y^2-4y=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\\ =-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)

đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\2y+1=0\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-0,5\end{matrix}\right.\)

vậy MAX A=7 tại \(\left\{{}\begin{matrix}x=1\\y=-0,5\end{matrix}\right.\)

\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

đặt: \(t=x^2+5x\) khi đó:

\(D=\left(t-6\right)\left(t+6\right)\\ D=t^2-36\ge-36\)

đẳng thức xảy ra khi :

\(t=0\\ \Leftrightarrow x^2+5x=0\\ x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

vậy MAX D=-36 tại x=0 hoặc x=-5