Question

已知xy都是正实数且满足4x²+4xy+y²+2x+y-6=0则x（1-y）的最小值

Answer 1

4x²+4xy+y²+2x+y-6=0
(2x+y)²+(2x+y)-6=0
(2x+y+3)(2x+y-2)=0
2x+y+3=0或2x+y-2=0
y=-2x-3或y=2-2x
x(1-y)=x(1+2x+3)
=2x²+4x
=2(x²+2x+1)-2
=2(x+1)²-2
≥-2
∴当x=-1, y=-1时则x（1-y）的最小值是-2
x(1-y)=x(1+2x-2)
=2x²-x
=2(x²-½x+¼)-½
=2(x-½﹚²-½
≥-½
∴当x=½, y=1时则x（1-y）的最小值是-½

Answer 2

4x²+4xy+y²+2x+y-6=0得2x+y=2
所以y=2-2x 带入x（1-y）求解 注意xy都是正实数这个条件

Answer 3

4x²+4xy+y²+2x+y-6=0
(2x+y)²+(2x+y)-6=0
(2x+y+3)(2x+y-2)=0
2x+y+3=0或2x+y-2=0
y=-2x-3或y=2-2x
x(1-y)=x(1+2x+3)
=2x²+4x
=2(x²+2x+1)-2
=2(x+1)²-2
≥-2
∴当x=-1,
y=-1时则x(1-y)的最小值是-2
x(1-y)=x(1+2x-2)
=2x²-x
=2(x²-½x+¼)-½
=2(x-½﹚²-½
≥-½
∴当x=½,
y=1时则x(1-y)的最小值是-½