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\(=\dfrac{3\cdot7\cdot3^4\cdot3^6+3^6\cdot3^4\cdot3^3}{3^2\cdot3^4\cdot2\cdot3^{12}\cdot13+3^2\cdot2\cdot3^3\cdot2\cdot3^4\cdot2\cdot3^2+723\cdot729}\)
\(=\dfrac{3^{11}\cdot7+3^{13}}{3^{18}\cdot26+3^{11}\cdot8+3^7\cdot241}\)
\(=\dfrac{3^{11}\left(7+9\right)}{3^7\left(3^{11}\cdot26+3^4\cdot8+241\right)}=\dfrac{3^7\cdot16}{17\cdot101\cdot2683}\)
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
Ta có :C= 2181-729+243.81-27
=2052+19683-27
C=21108
D=\(3^2.9^2.243+18.243.324.243\)
=9.81.243+18.243.324.243
=177147+344373768
=344550915
Ta có : C:D=21108:344550915=0,00006
729 > 3n ≥ 9.27
729 = 36 ; 9.27 = 243 = 35
=> 36 > 35 ≥ 35 hay 729 > 35 ≥ 9.27
Vậy n = 5