4x^2+4x+y^2-8y+17=0,求2x+y的值

题目

4x^2+4x+y^2-8y+17=0,求2x+y的值
试说明无论x,y为任何实数,代数式（x+y）^2-2x-2y+2的值都不会小於1

答案

4x^2+4x+y^2-8y+17=0,4x^2+4x+1+y^2-8y+16=0,(2x+1)^2+(y-4)^2=0(2x+1)^2=0,(y-4)^2=0x=-1/2,y=42x+y=(-1/2)*2+4=-1+4=3（x+y）^2-2x-2y+2=（x+y）^2-2(x+y)+2=（x+y）^2-2(x+y)+1+1=(x+y-1)^2+1因为(x+y-1)^2>=0所...