浙教版七年级下数学竞赛题一道

Question

求证：不论x、y为何值，多项式x^2-xy+y^2-2x+y+5/2的值永远大于或等于0.请写出解题过程，谢谢！

Answer 1

x^2-xy+y^2-2x+y+5/2x-y)=(1/2){x^2-2xy+x^2+y^2+y^2-4x+2y+5 }
=(1/2){(x-y)^2+x^2-4x+4-4+y^2+2y+5 }
=(1/2){(x-y)^2+(x-2)^2+y^2+2y+1 }
=(1/2){(x-y)^2+(x-2)^2+(y+1)^2 }>=0
得证

Answer 2

x^2-xy+y^2-2x+y+5/2=x^2/2+x^2/2-xy+y^2/2+y^2/2-2x+y+2+1/2=(x^2/2-xy+y^2/2)+(x^2/2-2x+2)+(y^2/2+y+1/2)=(x^2-2xy+y^2)/2+(x^2-4x+4)/2+(y^2+2y+1)/2=(x-y)^2/2+(x-2)^2/2+(y+1)^2/2[(x-y)^2+(x-2)^2+(y+1)^2]/2>=0

Answer 3

先原式*22*（x^2-xy+y^2-2x+y+5/2）=2x^2-2xy+2y^2-4x+2y+5=(x^2-2xy+y^2)+(x^2-4x+4)+(y^2+2y+1)=(x-y)^2+(x-2)^2+(y+1)^2∴原式=(x-y)^2+(x-2)^2+(y+1)^2/2∴不论x、y为何值,原式的值永远大于或等于0