因为4x^2+y^2-4x+2y+3=(4x^2-4x+1)+(y^2+2y+1)+1=(2x-1)^2+(y+1)^2+1,又因为(2x-1)^2≥0,(y+1)^2≥0,所以(2x-1)^2+(y+1)^2+1≥1,所以当x=1/2,y=-1时,代数式4x^2+y^2-4x+2y+3的值最小,这个最小值是1....