\(x+xy+y=35\)
\(\Rightarrow x+xy+y+1=36\)
\(\Rightarrow x\left(y+1\right)+\left(y+1\right)=36\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=36\)
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 \(\Leftrightarrow x+y+1=0\)