Question

[Châu Khánh Long]

chung minh khong ton tai 2 so x^2+2y^2-2xy+x-2y+1=0

Answer 1

[Rồng Đom Đóm]

\(x^2+2y^2-2xy+x-2y+1=0\)

\(\Leftrightarrow x^2-2xy+y^2+x-y+\dfrac{1}{4}+y^2-y+\dfrac{1}{4}+\dfrac{1}{2}=0\)

\(\Leftrightarrow\left(x-y\right)^2+2.\dfrac{1}{2}\left(x-y\right)+\left(\dfrac{1}{2}\right)^2+y^2-2.\dfrac{1}{2}y+\left(\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\)

\(\Leftrightarrow\left(x-y+\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\)

Mà \(VT\ge\dfrac{1}{2}>0\Rightarrow VT>VP\)

\(\Rightarrow\)PT vô nghiệm(đpcm)