x×x+y*y-2x+2y+40=x^2+y^2-2x+2y+40=x^2-2x+1+y^2+2y+1+38=(x-1)^2+(y+1)^2+38因为(x-1)^2>=0(y+1)^2>=0所以(x-1)^2+(y+1)^2+38>=38所以x×x+y*y-2x+2y+40的值恒为正数注意:^2表示平方.