có: p+ n+ e= 40
mà p= e nên: 2p+ n= 40
\(\Rightarrow\) n= 40- 2p (1)
mặt khác ta có: 1\(\le\) \(\frac{n}{p}\)\(\le\) 1,5
\(\Rightarrow\) p\(\le\) n\(\le\) 1,5p (2)
thay (1) vào(2) có:
p\(\le\) 40- 2p\(\le\) 1,5p
\(\Rightarrow\) \(\left\{{}\begin{matrix}p\text{}\text{}\text{}\text{}\text{}\text{}\le40-2p\\40-2p\le1,5p\end{matrix}\right.\)
\(\Rightarrow\) \(\frac{80}{7}\)\(\le\) p\(\le\) \(\frac{40}{3}\)
\(\Rightarrow\) \(\left[{}\begin{matrix}p=12\left(loại\right)\\p=13\left(nhận\right)\end{matrix}\right.\)
\(\Rightarrow\) e= p= 13; n= 14
kí hiệu: 2713X