Ta có

2B = 2x² + 2y² + 2xy - 6x - 6y + 4026

= (x² + 2xy + y²) - (4x + 4y) + (x² - 2x + 1) + (y² - 2y + 1) + 4 + 4020

= (x + y)² - 4(x + y) + 4 + (x - 1)² + (y - 1)² + 4020

= (x + y -2)² + (x - 1)² + (y - 1)² + 4020 \(\ge4020\)

=> B\(\ge2010\)

Đạt được khi x = y = 1

Gọi \(A=x^2+y^2+xy-3x-3y-3\)

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

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

\(=\left(x-1\right)^2+2\left(x-1\right)\frac{1}{2}\left(y-1\right)+\frac{1}{4}\left(y-1\right)^2+\frac{3}{4}\left(y-1\right)^2-6\)

\(=\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2+\frac{3}{4}\left(y-1\right)^2-6\ge-6\)Có GTNN là -6

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2=0\\\frac{3}{4}\left(y-1\right)^2=0\end{cases}\Rightarrow x=y=1}\)

Vậy GTNN của A là -6 tại x = y = 1

A= x²+y²+xy-3x-3y-3

\(=\left[x-1+\frac{1}{2}\left(y-1\right)\right]^2+\frac{3}{4}\left(y-1\right)^2-6\ge-6\)

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

Vậy.............

bạn tham khảo đi Tìm GTNH: P=x^2+xy+y^2-3x-3y+2010? | Yahoo Hỏi & Đáp

\(M=x^2+y^2+xy-3x-3y+2018\)

\(=x^2+2x\frac{\left(y-3\right)}{2}+\left(\frac{y-3}{2}\right)^2+y^2-3y+2018-\left(\frac{y-3}{2}\right)^2\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3y^2-6y+8063}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y^2-2y+1\right)}{4}+2015\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+2015\ge2015\)

\("="\Leftrightarrow x=y=1\)

Cảm ơn bạn nhiều nha

CHÚC BẠN HỌC TỐT

Thanks

Đặt biểu thức là A

\(x^2+xy+y^2-3x-3y+2018\)

\(=\left(x^2+xy+y^2\right)-\left(3x+3y\right)+2018\)

\(=\left(x+y\right)^2-3\left(x+y\right)+2018\)

Ta có : (x - y)² ≥ 0 
<=> x² + y² ≥ 2xy 
<=> x² + 2xy + y² ≥ 4xy 
<=> (x + y)² ≥ 4xy 
<=> xy ≤ (x + y)²/4 
<=> -xy ≥ -(x + y)²/4 

--> A ≥ (x + y)² - 3(x + y) - (x + y)²/4 

<=> A ≥ 3(x + y)²/4 - 3(x + y) 

để dễ nhìn,ta đặt t = x + y 

--> A ≥ 3t²/4 - 3t = 3(t²/4 - 2.t/2 + 1) - 3 = 3(t/2 - 1)² - 3 ≥ -3 

Dấu " = " xảy ra <=> t/2 = 1 <=> t = 2 <=> x + y = 2 và x = y --> x = y = 1 

Vậy MinA = -3 <=> x = y = 1

\(P=x^2+3x+y^2+3y+\frac{9}{x^2+y^2+1}\)

\(=x^2+y^2+1+\frac{9}{x^2+y^2+1}+3x+3y-1\)

\(\ge2.3.\frac{\sqrt{x^2+y^2+1}}{\sqrt{x^2+y^2+1}}+2.3.\sqrt{xy}-1\)

\(=6+6-1=11\)

Dấu = xảy ra khi x = y = 1