\(2x^2+y^2+2xy-6x-2y+10\)

\(=\left(x^2-4x+4\right)+\left(x^2+y^2+1+2xy-2y-2x\right)+5\)

\(=\left(x-2\right)^2+\left(x+y-1\right)^2+5\ge5\)

GTNN là 2015 nha  bạn

\(B=2x^2+y^2+2xy+6x+2y+2015\)

\(=x^2+y^2+1+2xy+2y+2x+x^2+4x+4+2011\)

\(=\left(x^2+y^2+1+2xy+2y+2x\right)+\left(x^2+4x+4\right)+2011\)

\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2011\)

Vì \(\left(x+y+1\right)^2+\left(x+2\right)^2\ge0\)nên \(\left(x+y+1\right)^2+\left(x+2\right)^2+2011\ge2011\)

Vậy \(MinB=2011\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x+2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

Min là giá trị nhỏ nhất mà, không biết àk

 P=2x²+y²-2xy-6x+2y+2024

=>2P=4x²+2y²-4xy-12x+4y+4048

=(2x-y-3)²+y²-2y+1+4038

=(2x-y-3)²+(y-1)²+4038> hoặc = 4038

Dấu = xảy ra <=>2x-y-3=0 và y-1=0=>x=2;y=1=>2p=4038=>p=2019

Vậy Pmin=2019<=>x=2;y=1

Ta có: 

P = 2x² + y² - 2xy - 6x + 2y + 2024

P = (x² - 2xy + y²) - 2(x - y) + 1 + (x² - 4x + 4) + 2019

P = [(x - y)² - 2(x - y) + 1] + (x - 2)² + 2019

P = (x - y - 1)² + (x - 2)² + 2019 \(\ge\)2019 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}}\) <=> \(\hept{\begin{cases}y=x-1\\x=2\end{cases}}\) <=> \(\hept{\begin{cases}y=1\\x=2\end{cases}}\)

Vậy MinP = 2019 <=> x = 2 và y = 1

\(P=2x^2+y^2-2xy-6x+2y+2024\)

\(\Rightarrow P=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+\left(x^2-4x+4\right)+2019\)

\(\Rightarrow P=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+\left(x-2\right)^2+2019\)

\(\Rightarrow P=\left(x-y-1\right)^2+\left(x-2\right)^2+2019\)

Ta có:

\(\left(x-y-1\right)^2\ge0\forall x;y\inℝ\)

\(\left(x-2\right)^2\ge0\forall x\inℝ\)

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

\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2+2019\ge2019\forall x;y\inℝ\)

\(\Rightarrow P\ge2019\forall x;y\inℝ\)

Dấu "=" xảy ra:

\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-1\\x=2\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=2\end{cases}}}\)

Vậy P nhỏ nhất khi P = 2019 tại x=2;y=1

Chúc bạn học tốt nhé!

2x² + y² + 2xy - 6x - 2y + 10

= x² + y² + 1² + 2xy - 2x - 2y + x² - 4x + 4 + 5

= (x + y - 1)² + (x - 2)² + 5  5

Dấu ''='' xảy ra khi {x+y−1=0x−2=0{x+y−1=0x−2=0 ⇔{y=−1x=2⇔{y=−1x=2

Vậy Min = 5 khi x = 2 và y = - 1

Ta có: \(B=2x^2+y^2-2xy+6x+10\)

\(=x^2-2xy+y^2+x^2+6x+9+1\)

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

Dấu '=' xảy ra khi x=y=-3

Vậy: \(B_{min}=1\) khi (x,y)=(-3;-3)

2A = 4x² + 4xy + 4y² - 12x - 12y + 8040

= (2x + y)² - 6(2x + y) + 9 + 3y² - 6y + 3 + 8028

= (2x + y - 3)² + 3(y - 1)² + 8028 \(\ge8028\)

=> \(A\ge4014\)

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

Vậy Min A = 4014 khi x = y = 1

Ta có: \(E=2x^2+2x\left(y+3\right)+2y^2+2020\)

\(=2\left(x^2+2.x.\frac{\left(y+3\right)}{2}+\frac{\left(y+3\right)^2}{4}\right)+2y^2+2020-\frac{\left(y+3\right)^2}{2}\)

\(=2\left(x+\frac{y+3}{2}\right)^2+\frac{3y^2-6y+4031}{2}\)

\(=2\left(x+\frac{y+3}{2}\right)^2+\frac{3\left(y-1\right)^2+4028}{2}\ge\frac{4028}{2}=2014\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-\frac{y+3}{2}\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

Vậy...

\(B=9x^2-6x+5=9x^2-6x+1+4\\ B=\left(3x-1\right)^2+4\ge4\)

đẳng thức xảy ra khi 3x-1=0 => x=1/3

vậy min B=4 tại x=1/3

\(C=x^2+x-3\)

\(C=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{2}^2-\dfrac{1}{2}^2-3\)

\(C=\left(x+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\)

Ta có: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow C\ge-\dfrac{13}{4}\)

Vậy Min_C=-13/4 khi x=-1/2

\(D=2x^2+2xy+y^2-2x+2y+2\)

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

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

Min_D=-3 khi x=2; y=-3

B=9x²-6x+5

=9

= 9\(\left(x-\dfrac{1}{3}\right)^2+4\)

(x²+2xy+y²)-2(x+y)+1+(x²-4x+4)+5

=(x+y-1)²+(x-2)²+5>=5

Bạn ơi đề k sai đúng hông?

2x^2+y^2+2xy-6x-2y+10

=(x+y+1)^2+(x-2)^2+5>=5

Amin=5 khi : x+y+1=0

               hoac  x-2=0

=>x=2 va y=-3