\(x+xy+y=35\)

\(\Rightarrow x+xy+y+1=36\)

\(\Rightarrow x\left(y+1\right)+\left(y+1\right)=36\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=36\)

Theo Cô si ta có:

\(36=\left(x+1\right)\left(y+1\right)\le\dfrac{\left[\left(x+1\right)+\left(y+1\right)\right]^2}{4}=\dfrac{\left(x+y+2\right)^2}{4}\)

\(\Rightarrow\left(x+y+2\right)^2\ge144\)

\(\Rightarrow x+y+2\ge12\)

\(\Rightarrow x+y\ge10\)

Lại có: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\ge10^2=100\)

\(\Rightarrow x^2+y^2\ge50\)

Dấu "=" xảy ra khi \(x=y=5\)

Cho em hỏi cách giải sau sai ở đâu ạ :(

\(x+y+xy=35\)

\(\Leftrightarrow2x+2y+2xy=70\)

\(\Leftrightarrow2xy=70-2x-2y\)

Mặt khác ta có :

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

\(=\left(x+y\right)^2-70+2x+2y\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1-71\)

\(=\left(x+y+1\right)^2-71\ge-71\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow x+y+1=0\)

2/xy<=1/x^2+1/y^2=1/2

=>xy>=4

Dấu = xảy ra khi x=y=2

(x+y)^2>=4xy>=16

=>x+y>=4

Dấu = xảy ra khi x=y=2

=>x+y+xy+2023>=2023+4+4=2031 

Dấu = xảy ra khi x=y=2

Đặt A = x^2+y^2-xy-x-y+2

4A = 4x^2+4y^2-4xy-4x-4y+8

= [(4x^2-4xy+y^2)-(4x-2y)+1]+(3y^2-6y+3)+4

= [(2x-y)^2-2.(2x-y)+1]+3.(y^2-2y+1)+4

= (2x-y+1)^2+3.(y-1)^2+4 >= 4 => A >= 1

Dấu "=" xảy ra <=> 2x-y-1=0 và y-1=0 <=> x=y=1

Vậy GTNN của A = 1 <=> x=y=1

k mk nha

Đặt \(B=x^2+y^2-xy-x-y+2\)

\(\Rightarrow4B=4x^2+4y^2-4xy-4x-4y+8\)

\(=\left[\left(4x^2+4xy+y^2\right)-2\left(2x+y\right)+1\right]+3\left(y^2-2y+1\right)+4\)

\(=\left[\left(2x+y\right)^2-2\left(2x+y\right)+1^2\right]+3\left(y-1\right)^2+4\)

\(=\left(2x+y-1\right)^2+3\left(y-1\right)^2+4\ge4\)

Dấu bằng khi x = 0, y = 1