x2+y2-xy-x-y+2 Tìm GTTN

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 = 1Câu trả lời: 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