ta có D=x^2 +2.y^2 -2xy+4x-5y-12

<=>D=(x^2 +y^2 +4 -2xy-4y+4x) +[y^2 -2.y.(1/2) +1/4] -1/4+8

<=>D=(x-y+2)^2 +(y-1/2)^2  +31/4

mà (x-y+2)^2 >= 0 và (y-1/2)^2>=0 nên (x-y+2)^2 +(y-1/2)^2 +31/4 >= 31/4

dấu '=' xảy ra khi :y-1/2=0 và x-y+2=0 <=> y=1/2 và x=-3/2

vậy GTNN của D là 31/4 khi x=-3/2, y=1/2

a) -( x-y)² - (x-1)² -2 

GTLN = -2

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)

\(H=x^2+2xy+y^2+2x+2y+x^2+4x+2019=\left(x+y\right)^2+2\left(x+y\right)+\left(x+2\right)^2+2015\)

\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2014\ge2014\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2;y=1\)

\(I=\left(1-x\right)^2+\left(-2-y\right)^2+\left(x+y\right)^2\ge\frac{\left(1-x-2-y+x+y\right)^2}{3}=\frac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(1-x=-2-y=x+y\)\(\Leftrightarrow\)\(x=\frac{4}{3};y=\frac{-5}{3}\)

E = 2x^2 - 5x -2 = 2( x^2 -5/2x -1) = 2(x^2 - 2.x.5/4 +25/16 - 41/16) = 2(x - 5/4 )^2 + 41/8

Vậy GTNN của biểu thức là 41/8 tại x = 5/4

F = x^2 + 5y^2 + 2xy -y +3 = (x^2 + 2xy +y^2) + (4y^2 - 2.2y.1/4 + 1/16) +47/16

(x + y)^2  + (2y - 1/4)^2 + 47/16 

Vậy GTNN của BT là 47/16 tại x = y = 1/8

đề bài sai r bn ơi phải là +10 chứ ko phải +8 đâu nhá

1. \(A=2x^2-6x-2xy+y^2+10\)

\(\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1\)

\(\Leftrightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\)

Vì \(\left(x-y\right)^2\ge0\) ; \(\left(x-3\right)^2\ge0\)\(\forall x;y\)

\(\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=3\)

Vậy minA = 1 \(\Leftrightarrow x=y=3\)

2. \(A=5+2xy+14y-x^2-5y^2-2x\)

\(\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15\)

\(\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\)

Vì \(\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\\left(2y-3\right)^2\ge0\end{matrix}\right.\)\(\forall x;y\)

\(\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\y=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

Vậy maxA = 15 \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

1.

Vì ;

Dấu "=" xảy ra

Vậy minA = 1

2.

Vì

Dấu "=" xảy ra

Vậy maxA = 15

\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)