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\(A=\frac{2^{19}3^9+5.2^{18}3^9}{3^9.2^9.2^{10}+3^{10}4^{10}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^92^{19}+3^{10}2^{20}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^9\left(2^{19}+3.2^{20}\right)}=\frac{2^{19}+5.2^{18}}{2^{19}+3.2^{20}}=\frac{2^{18}\left(2+5\right)}{2^{18}\left(2+3.2^2\right)}\)
\(=\frac{2+5}{2+3.2^2}=\frac{7}{14}=\frac{1}{2}\)
Có: \(5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}=5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}\)
\(=3^{18}\cdot2^{29}\cdot\left(5\cdot2-3^2\right)=3^{18}\cdot2^{29}\)
Lại có: \(5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6=5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}=5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(=2^{19}\cdot3^{18}\cdot\left(5\cdot3-7\right)=2^{19}\cdot3^{18}\cdot2^3=2^{22}\cdot3^{18}\)
Vậy \(\frac{3^{18}\cdot2^{29}}{2^{22}\cdot3^{18}}=2^7=128\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{19}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6}{7}\)