=x²-2xy+y²+4y²+4y+1+2

=(x-y)²+(2y+1)²+2\(\ge2\)

dấu bằng xảy ra khi x=y=-1/2

a) A= 2x²-8x+10 = 2(x-2)²+2\(\ge\)2\(\Leftrightarrow\)x=2

Vậy MinA=2 \(\Leftrightarrow\)x=2

b) B= -(x-1)²-(2y+1)²+7 \(\le\)7

Dấu = xảy ra khi x=1 và y=\(\frac{-1}{2}\)

Vậy MaxB=7 ....

cảm ơn bạn nha

Giups mik vs

a)Ta có: \(A=x^2+5y^2-2xy+4y+3\)= \(\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

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

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

Vậy min A=2. Dấu = khi x=y=-1/2

b) Đặt \(t=x^2-2x+1\)

=> \(B=\left(t-1\right)\left(t+1\right)\)=\(t^2-1\)=\(t^2+\left(-1\right)\ge-1\)

Do \(t^2\ge0\)

Vậy min B=-1. Dấu = khi t=0 hay \(x^2-2x+1=0\)

                                          => \(\left(x-1\right)^2=0\)<=> x=1

trời ơi ghi cả 1 dãy 

sao vậy

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=x^2-8x+16+x^2+4xy+4y^2+y^2+4y+4+2004\)

\(=\left(x-4\right)^2+\left(x+2y\right)^2+\left(y+2\right)^2+2004\ge2004\)

Dấu ''='' xảy ra khi x = 4 ; y = -2 

a) -( x-y)² - (x-1)² -2 

GTLN = -2

a) Đặt \(A=x^2-2x+1\)

    Ta có: \(A=x^2-2x+1=\left(x-1\right)^2\)

     Vì \(\left(x-1\right)^2\ge0\forall x\)

    \(\Rightarrow A_{min}=0\)

    Dấu "=" xảy ra khi: \(x-1=0\)

                            \(\Leftrightarrow x=1\)

Vậy \(A_{min}=0\)\(\Leftrightarrow\)\(x=1\)

b) Ta có: \(M=x^2-3x+10\)

        \(\Leftrightarrow M=\left(x^2-3x+\frac{9}{4}\right)+\frac{31}{4}\)

        \(\Leftrightarrow M=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)

    Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\forall x\)

     \(\Rightarrow\)\(M_{min}=\frac{31}{4}\)

    Dấu "=" xảy ra khi: \(x-\frac{3}{2}=0\)

                            \(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(M_{min}=\frac{31}{4}\)\(\Leftrightarrow\)\(x=\frac{3}{2}\)

Mình đặt A, B, C cho dễ nhìn nhé ;-;

a) A = x² - 2x + 1 = ( x - 1 )² ≥ 0 ∀ x

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinA = 0 <=> x = 1

b) B = x² - 3x + 10 = ( x² - 3x + 9/4 ) + 31/4 = ( x - 3/2 )² + 31/4 ≥ 31/4 ∀ x

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MinB = 31/4 <=> x = 3/2

c) C = ( x - 3 )( x + 5 ) + 4

= x² + 2x - 15 + 4

= ( x² + 2x + 1 ) - 12

= ( x + 1 )² - 12 ≥ -12 ∀ x

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MinC = -12 <=> x = -1

d) D = x² - 4x + y² - 8y + 6

= ( x² - 4x + 4 ) + ( y² - 8y + 16 ) - 14

= ( x - 2 )² + ( y - 4 )² - 14 ≥ -14 ∀ x, y

Đẳng thức xảy ra <=> x = 2 ; y = 4

=> MinD = -14 <=> x = 2 ; y = 4

e) E =  3x² + 2x + 1

= 3( x² + 2/3x + 1/9 ) + 2/3

= 3( x + 1/3 )² + 2/3 ≥ 2/3 ∀ x

Đẳng thức xảy ra <=> x + 1/3 = 0 => x = -1/3

=> MinE = 2/3 <=> x = -1/3

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