Ta có:
\(\left\{{}\begin{matrix}p_A=n_A\\p_B-n_B=1\end{matrix}\right.\left(\text{*}\right)\)
\(p_A+4p_B=10\left(\text{*}\text{*}\right)\Leftrightarrow\frac{p_A+n_A}{p_A+n_A+4p_B+4n_B}=0,75\)
\(\Leftrightarrow p_A+n_A=0,75.\left(p_A+n_A+4p_B+4n_B\right)\left(\text{*}\text{*}\text{*}\right)\)
\(\left(\text{*}\right)+\left(\text{*}\text{*}\text{*}\right)\Rightarrow2p_A=0,75.\left(2p_A+4p_B+4p_B-4\right)\)
\(\Leftrightarrow0,5p_A-6p_B=-3\left(\text{*}\text{*}\text{*}\text{*}\right)\)
\(\left(\text{*}\text{*}\right)+\left(\text{*}\text{*}\text{*}\text{*}\right)\Rightarrow\left\{{}\begin{matrix}p_A=6\\p_B=1\end{matrix}\right.\)
Vậy A là C, B là H.