\(A=-x^2+2xy-4y^2+2x+10y-3\)

\(=-x^2+2xy-y^2+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10< =10\)

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

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

a) -( x-y)² - (x-1)² -2 

GTLN = -2

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(E=2x^2+5y^2+x+4y+5\)

\(\Rightarrow E=2x^2+x+5y^2+4y+5\)

\(\Rightarrow E=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}-\dfrac{1}{16}\right)+5\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}-\dfrac{4}{25}\right)+5\)

\(\Rightarrow E=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+5\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)+5-\dfrac{1}{8}-\dfrac{4}{5}\)

\(\Rightarrow E=2\left(x+\dfrac{1}{4}\right)^2+5\left(y+\dfrac{2}{5}\right)^2+\dfrac{163}{40}\)

mà \(\left\{{}\begin{matrix}2\left(x+\dfrac{1}{4}\right)^2\ge0,\forall x\\5\left(y+\dfrac{2}{5}\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow E=2\left(x+\dfrac{1}{4}\right)^2+5\left(y+\dfrac{2}{5}\right)^2+\dfrac{163}{40}\ge\dfrac{163}{40}\)

\(\Rightarrow GTNN\left(E\right)=\dfrac{163}{40}\left(tạix=-\dfrac{1}{4};y=-\dfrac{2}{5}\right)\)

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

\(=x^2-4xy+4y^2+x^2-2x+1+2012\)

\(=\left(x-2y\right)^2+\left(x-1\right)^2+2012\ge2012\)

Dấu = xảy ra khi : \(\left(x-1\right)^2=0\Leftrightarrow x=1\)

                              \(\left(x-2y\right)^2=0\Leftrightarrow2y=1\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(Min_B=2012\) khi x=1 , y=\(\dfrac{1}{2}\)

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Ta có:

\(\left(x-1\right)^2+\left(y+2\right)^2=0\)

Do: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

Mặt khác: \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Thay vào B ta có:

\(B=2\cdot1^5-5\cdot\left(-2\right)^3+4=2\cdot1-5\cdot-8+4=2+40+4=46\)

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