Lời giải:
\(\frac{5.(2^2)^{15}.(3^2)^9-2^2.3^{20}.(2^3)^9}{5.2^9.2^{19}.3^{19}-7.2^{29}.(3^3)^6}\\ =\frac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\\ =\frac{2^{29}.3^{18}(2.5-3^2)}{2^{28}.3^{18}(5.3-7.2)}=\frac{2^{29}.3^{18}}{2^{28}.3^{18}}=2\)