说理:试说明不论x,y取什么有理数,多项式x^2+y^2-2x+2y+3的值总是正数

x^2+y^2-2x+2y+3
=(x^2-2x+1)+(y^2+2y+1)+1
=(x-1)^2+(y+1)^2+1
因为平方非负
所以(x-1)^2+(y+1)^2>=0
所以(x-1)^2+(y+1)^2+1>0,x^2+y^2-2x+2y+3>0

x^2+y^2-2x+2y+3
=x²-2x+1+y²+2y+1+1
=(x-1)²+(y+1)²+1
∵(x-1)²≥0 (y+1)²≥0
∴(x-1)²+(y+1)²+1＞0
∴不论x,y取什么有理数,多项式x^2+y^2-2x+2y+3的值总是正数

x^2+y^2-2x+2y+3=(x^-2x+1)+(y^2+2y+1)+1=(x-1)^2+(y+1)^2+1 当x=1,y=-1时。值最小=1