Ta có: \(\left\{{}\begin{matrix}p-n=1\\p+e-n=11\\p=e\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p-n=1\\2p-n=11\\p=e\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=10\\p-n=1\\p=e\end{matrix}\right.\left\{{}\begin{matrix}p=e=10\\n=9\end{matrix}\right.\)
Đúng 2
Bình luận (0)