Question

求代数式x²-xy+y²-2x+y的最小值

Answer 1

x²-xy+y²-2x+y 将它化做

= (2x²-2xy+2y²-4x+2y)/2

= (x²-2xy+y²+x²-4x+4-4+y²+2y+1-1)/2

= ((x+y)^2+(x-2)^2+(y+1)^2-5)/2

使任意两平方内的数为0,则可得到最小值 -2

如x+y=0 x-2=0 x=2 y=-2

((-2+1)^2-5)/2=-2

x+y=0 y+1=0 y=-1 x=1

((1-2)^2-5)/2=-2

x-2=0 y+1=0 x=2 y=-1

((-1+2)^2-5)/2=-2