Lời giải:
Gọi phân số trên là $P$
\(P=\frac{(2^2)^{10}.(3^2)^6+3^{12}.(2^3)^5}{2^{13}.3^{13}.2^2-2^{16}.3^{12}}\) \(=\frac{2^{20}.3^{12}+3^{12}.2^{15}}{2^{15}.3^{13}-2^{16}.3^{12}}=\frac{3^{12}.2^{15}(2^5+1)}{2^{15}.3^{12}(3-2)}=\frac{2^{5}+1}{1}=33\)