\(=\frac{5\cdot2^{20}\cdot3^{11}+7\cdot2^{29}\cdot3^{18}}{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}=\frac{2^{20}\cdot3^{11}\left(5+7\cdot2^9\cdot3^7\right)}{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}=\frac{5+7\cdot2^9\cdot3^7}{2^9\cdot3^7\cdot1}\)

...

\(=\dfrac{5\cdot2^{30}-3^{20}\cdot2^{29}}{5\cdot2^{10}-7\cdot2^{30}\cdot3^3}\)

\(=\dfrac{2^{29}\left(5\cdot2-3^{20}\right)}{2^{10}\left(5\cdot1-7\cdot2^{20}\cdot3^3\right)}=\dfrac{2^{19}\cdot\left(10-3^{20}\right)}{5-189\cdot2^{20}}\)

......

\(=\dfrac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{13}-3^{29}\cdot5^{19}}=\dfrac{3^{29}\cdot5^{20}-5^{18}\cdot3^{31}}{3^{29}\cdot5^{13}\cdot7-3^{29}\cdot5^{19}}\)

\(=\dfrac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{13}\left(7-5^6\right)}=5^5\cdot\dfrac{4^2}{7-5^6}\)

1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)

\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)

\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)

B={x\(\in\)N|x=3^k; 1<=k<=4}

C={x\(\in\)N|x=4*a²; 1<=a<=5}

D={x\(\in\)N|x=9*a²;1<=a<=4}

E={x\(\in\)N|x=4k; 0<=x<=4}

G={x\(\in\)N|x=(-3)^k; 1<=k<=4}

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)