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Ta có: \(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}\left(5\cdot2-3^2\right)}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}}{2^{29}\cdot3^{18}\left(5\cdot2-7\right)}\)
\(=\dfrac{1}{3}\)
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
\(=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}+7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3+7.2\right)}\)
\(=\frac{2}{29}\)
a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)
b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)
c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)