a)Đặt A=\(x^2-4xy+5y^2-2y+3\)

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

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

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

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

Dấu = xảy ra khi \(\hept{\begin{cases}x-2y=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2y\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

          Vậy Min A = 2 khi x = 2 ; y = 1

b)k ko hỉu

a)A= \(x^2-4xy+5y^2-2y+3\)

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

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

MIN A=-2 khi\(\orbr{\begin{cases}x-2y=0\\y-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\y=1\end{cases}}}\)Vậy.......

b)\(B=x^2-2xy+2y^2-x+y\)????

6 nha bạn

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

=(x^2+y^2+2xy​)+(2x+2y)+3

=((x+y)² +2(x+y) +1)+2

=(x+y+1)²+2

vậy A_min=2

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

<=>\(A=x^2+2x\left(y+1\right)+y^2+2y+3\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y^2+2y+1\right)+2\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2+2\)

<=>\(A=\left(x+y+1\right)^2+2\ge2\)