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\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^9}{2^9.3^9.2^{10}+\left(3.2^2\right)^{10}}\)
\(=\frac{2^{19}.3^9+5.2^{18}.3^{19}}{2^{19}.3^9+3^{10}.2^{20}}=\frac{2^{18}.3^9\left(2+5.3\right)}{2^{18}.3^9\left(2+3.2^2\right)}=\frac{17}{14}\)
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\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)
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Ta có: \(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}\left(5\cdot2-3^2\right)}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}}{2^{29}\cdot3^{18}\left(5\cdot2-7\right)}\)
\(=\dfrac{1}{3}\)
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\(A=\frac{2^{19}3^9+5.2^{18}3^9}{3^9.2^9.2^{10}+3^{10}4^{10}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^92^{19}+3^{10}2^{20}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^9\left(2^{19}+3.2^{20}\right)}=\frac{2^{19}+5.2^{18}}{2^{19}+3.2^{20}}=\frac{2^{18}\left(2+5\right)}{2^{18}\left(2+3.2^2\right)}\)
\(=\frac{2+5}{2+3.2^2}=\frac{7}{14}=\frac{1}{2}\)
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