Question

当x,y各为多少时（x+2y）²-2xy-2x-2y+8的值 最小 求详细过程

Answer 1

x^2+4xy+4y^2-2xy-2x-2y+8
=x^2+2xy+4y^2-2x-2y+8
=(x^2+2xy+y^2)+3y^2-2x-2y+8
=(x+y)^2+3y^2-2(x+y)+1+7
=[(x+y)^2-2(x+y)+1]+3y^2+7
=[(x+y)-1]^2+3y^2+7
最小值为7，所以同时满足y=0，x+y-1=0
3y^2=0 ∴y=0
x+y-1=0 所以x=1

Answer 2

（x+2y）²-2xy-2x-2y+8
=x^2+4xy+4y^2-2xy-2x-2y+8
=x^2+2xy+4y^2-2x-2y+8
=(x+y)^2+3y^2-2(x+y)+1+7
=[(x+y)^2-(x+y)+1]+3y^2+7
=(x+y-1)^2+3y^2+7
该式最小值为7，即同时满足y=0，x+y-1=0是上式大于等于7，所以x=1，y=0

Answer 3

x^2+4xy+4y^2-2xy-2x-2y+8
=x^2+2xy+4y^2-2x-2y+8
=(x^2+2xy+y^2)+3y^2-2x-2y+8
=(x+y)^2+3y^2-2(x+y)+1+7
=[(x+y)^2-2(x+y)+1]+3y^2+7
=[(x+y)-1]^2+3y^2+7
因为除7外的两个式子都大于等于0，所以当两式同为0时有最小值为7
3y^2=0 ;y=0
x+y-1=0 ;x=1
写的详细，自己简化下