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Last edited:
Cựu Mod Toán
Thành viên
TV BQT tích cực 2017
Đặt $2^{2018}=a$, $3^{2019}=b$, $5^{2020}=c$ $(a,b,c>0)$
$\Rightarrow A=\dfrac a{a+b} + \dfrac b{b+c} + \dfrac c{c+a} > \dfrac a{a+b+c} + \dfrac b{a+b+c} + \dfrac c{a+b+c} = 1 > \dfrac{2019}{2020}$
--> Đề sai.^^