Cho a,b >0 thỏa mãn$a^2+b^2+ab\le 5-a-b$
Tìm min $F=\left(ab+2\right)\cdot \left(\frac{1}{a}+\frac{1}{b}\right)$
Tìm Min F=(ab+2)(1/a+1/b)
Started By YouKnow, 14-01-2019 - 20:51
#1
Posted 14-01-2019 - 20:51
YouKnow
-
- Thành viên mới
-
- 5 posts
Lính mới
- Diepnguyencva and buingoctu like this
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
