已知实数x,y满足x2-2xy+4y2-x-2y+1≤0,求x和y的值

题目

已知实数x,y满足x2-2xy+4y2-x-2y+1≤0,求x和y的值

答案

x2-2xy+4y2-x-2y+1≤0,
整理得x^2-(2y+1)x+4y^2-2y+1<=0,
配方得[x-(y+1/2)]^+3y^-3y+3/4<=0,
再配方得[x-(y+1/2)]^2+3(y-1/2)^2<=0,
x,y是实数,
∴x-(y+1/2)=0,y-1/2=0,
解得x=1,y=1/2.