ta có \(A=x^2+y^2+9-2xy-6x+6y+x^2-4x+4+2004\)

             \(=\left(x-y-3\right)^2+\left(x-2\right)^2+2004\)

vì \(\left(x-y-3\right)^2+\left(x-2\right)^2\ge0\)

=> \(A\ge2004\)

dấu = xảy ra <=> x=2 và y=-1

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=-2\)

\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)

Dấu \("="\Leftrightarrow x=-5\)

\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

\(=\left(x+2\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=-2

\(C=4x^2-4x+5\)

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

\(=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

A = x² - 10x + 12

= ( x² - 10x + 25 ) - 13

= ( x - 5 )² - 13

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 13 ≥ -13

Đẳng thức xảy ra <=> x - 5 = 0 => x = 5

=> MinA = -13 <=> x = 5

B = 6y² + 4y - 1

= 6( y² + 2/3y + 1/9 ) - 5/3

= 6( y + 1/3 )² - 5/3

6( y + 1/3 )² ≥ 0 ∀ x => 6( y + 1/3 )² - 5/3 ≥ -5/3

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

=> MinB = -5/3 <=> y = -1/3

C = x² + y² - 2x - 6y - 1

= ( x² - 2x + 1 ) + ( y² - 6y + 9 ) - 11

= ( x - 1 )² + ( y - 3 )² - 11

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y-3\right)^2-11\ge-11\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

=> MinC = -11 <=> x = 1 ; y = 3

D = 2x² + 3y² - x - 3y + 5

= 2( x² - 1/2x + 1/16 ) + 3( y² - y + 1/4 ) + 33/8

= 2( x - 1/4 )² + 3( y - 1/2 )² + 33/8

\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x-\frac{1}{4}\right)^2+3\left(y-\frac{1}{2}\right)^2+\frac{33}{8}\ge\frac{33}{8}\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{4}=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{1}{2}\end{cases}}\)

=> MinD = 33/8 <=> x = 1/4 ; y = 1/2

\(D=\left(x^2+y^2+1^2+2\left(x-y-xy\right)\right)+\left(y^2-4y+4\right)+\left(2020-1-16\right)\)\(D=\left(x-y+1\right)^2+\left(y-2\right)^2+2015\ge2015\)

https://olm.vn/hoi-dapDễ z mà ko bít ..

a) Ta có: 3x - x² = -(x² - 3x + 9/4) + 9/4 = -(x - 3/2)² + 9/4 \(\le\)9/4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2

Vậy Max của 3x - x² = 9/4 <=> x = 3/2

b) Ta có: x² - 6x + 18 = (x² - 6x + 9) + 9 = (x - 3)² + 9 \(\ge\)9 \(\forall\)x

Dấu "=" xảy ra <=>  x - 3 = 0 <=> x = 3

Vậy Min của x² - 6x + 18 = 9 <=> x = 3

c) Ta có : 2x² + 10x - 1 = 2(x² + 5x + 25/4) - 27/2 = 2(x + 5/2)² - 27/2 \(\ge\)-27/2 \(\forall\)x

Dấu "=" xảy ra <=> x + 5/2 = 0 <=> x = -5/2

Vậy Min của 2x² + 10x - 1 = -27/2 <=> x = -5/2

d) Ta có : x² + y² - 2x + 6y + 2019

= (x² - 2x + 1) + (y² + 6y + 9) + 2009

= (x - 1)² + (y + 3)² + 2009 \(\ge\)2009 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

Vậy Min của x² + y² - 2x + 6y + 2019 = 2009 <=> x = 1 và y=  -3