Question

已知实数(x^2+2x+3)(y^2+2/3x+1/3)=4/9求x+y的值

Answer 1

原方程化为
((x+1)^2+2)((y+1/3)^2+2/9)=4/9
又(x+1)^2≥0 (y+1/3)^2≥0
所以(x+1)^2+2≥2 (y+1/3)^2+2/9≥2/9
((x+1)^2+2)((y+1/3)^2+2/9)≥4/9
只有当(x+1)^2=0 (y+1/3)^2=0时
原方程成立
此时x=-1 y=-1/3
则x+y=-4/3

Answer 2

(x^2+2x+3)(y^2+2/3x+1/3)=4/9
(x^2+2x+1+2)(y^2+2/3x+1/9+2/9)=4/9
((x+1)^2+2)((y+1/3)^2+2/9)=4/9
又x,y是实数
(x+1)^2>=0
(y+1/3)^2>=0
左>=4/9=右
如果等式成立
必须有
(x+1)^2=0
(y+1/3)^2=0
即
x+1=0
y+1/3=0
解得
x=-1
y=-1/3
则
x+y=-4/3

Answer 3

解：
x^2+2x+3=(x+1)^2+2≥2，
y^2+2/3y+1/3=(y+1/3)^2+2/9≥2/9，
所以(x^2+2x+3)(y^2+2/3x+1/3)≥4/9，
仅当x=-1，y=-1/3时等号成立，
所以x+y=-1-1/3=-4/3.
O(∩_∩)O~

Answer 4

(x^2+2x+3)(y^2+2y/3+1/3)
=[(x+1)^2+2][(y+1/3)^2+2/9]
=4/9
(x+1)^2=0 (y+1/3)^2=0
x=-1,y=-1/3
x+y=-4/3

Answer 5

（X^2+2X+3）（Y^2+2Y/3+1/3)
=[(x+1)^2+2] [(y+1/3)^2 +2/9]
=（x+1)^2(y+1/3)^2+2/9(x+1)^2+2(y+1/3)^2+4/9=4/9
所以（x+1)^2(y+1/3)^2+2/9(x+1)^2+2(y+1/3)^2=0
所以x+1=0 y+1/3=0 即x+y=-4/3