Chủ đề này có 2 trả lời

[DOTOANNANG]

$\it{6}^{\,\it{4}}+ \it{5}^{\,\it{4}}+ \it{3}^{\,\it{4}}+ \it{2}^{\,\it{4}}+ \it{1}^{\,\it{4}}= \it{2019}$

 

$\it{2}^{\,\it{11}}- \it{2}^{\,\it{4}}- \it{2}^{\,\it{3}}- \it{2}^{\,\it{2}}- \it{2}^{\,\it{0}}= \it{2019}$

 

$\it{3}^{\,\it{7}}- \it{3}^{\,\it{5}}+ \it{3}^{\,\it{4}}- \it{3}^{\,\it{2}}+ \it{3}^{\,\it{1}}= \it{2019}$

 

$\it{5}^{\,\it{5}}- \it{4}^{\,\it{5}}- \it{3}^{\,\it{4}}- \it{1}^{\,\it{4}}= \it{2019}$

 

$\it{7}^{\,\it{4}}- \it{6}^{\,\it{3}}- \it{5}^{\,\it{3}}- \it{4}^{\,\it{3}}+ \it{3}^{\,\it{3}}- \it{2}^{\,\it{2}}= \it{2019}$

 

$\it{2}^{\,\it{10}}+ \it{2}^{\,\it{9}}+ \it{2}^{\,\it{8}}+ \it{2}^{\,\it{7}}+ \it{2}^{\,\it{6}}+ \it{2}^{\,\it{5}}+ \it{2}^{\,\it{4}}- \it{2}^{\,\it{3}}- \it{2}^{\,\it{2}}- \it{2}^{\,\it{1}}+ \it{2}^{\,\it{0}}= \it{2019}$

[DOTOANNANG]

$\it{201}\,.\,\it{9}+ \it{201}+ \it{9}= \it{2019}$

[DOTOANNANG]

$\lceil$ T ổ n g   b ì n h   p h ư ơ n g   c ủ a   $\it{3}$   s ố   n g u y ê n   t ố  ! $\rfloor$

 

$\it{2019}= \it{7}^{\,\it{2}}+ \it{11}^{\,\it{2}}+ \it{43}^{\,\it{2}}$

 

$\it{2019}= \it{7}^{\,\it{2}}+ \it{17}^{\,\it{2}}+ \it{41}^{\,\it{2}}$

 

$\it{2019}= \it{13}^{\,\it{2}}+ \it{13}^{\,\it{2}}+ \it{41}^{\,\it{2}}$

 

$\it{2019}= \it{11}^{\,\it{2}}+ \it{23}^{\,\it{2}}+ \it{37}^{\,\it{2}}$

 

$\it{2019}= \it{17}^{\,\it{2}}+ \it{19}^{\,\it{2}}+ \it{37}^{\,\it{2}}$

 

$\it{2019}= \it{23}^{\,\it{2}}+ \it{23}^{\,\it{2}}+ \it{31}^{\,\it{2}}$

 

Spoiler

$\it{2019}$  là số nguyên nhỏ nhất có thể viết dưới dạng tổng bình phương $\it{3}$ số nguyên tố theo $\it{6}$ cách !