c) G = \(\frac{636363.37-373737.63}{1+2+3+...+2017}\)

G = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2017}\)

G = \(\frac{0}{1+2+3+...+2017}\)

=> G = 0

Vậy G = 0

a) \(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}.\frac{612}{1225}\)

\(\Rightarrow E=\frac{306}{1225}\)

Vậy...

b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)

d) Bạn xem lại đề nhé

Câu a:

(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)

= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5

= 1 + 15 - 3 4/5

= 16 - 3 - 4/5

= 13 - 4/5

= 61/5

Câu b:

(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))

= (17/6 + 13/9) :(121/12 - 19/2)

= (51/18 + 26/18) :(121/12 - 114/2)

= 77/18 : 7/12

= 77/18 x 12/7

= 22/3

\(\frac{5}{23}x\frac{17}{6}+\frac{5}{23}x\frac{9}{26}\)

\(=\frac{5}{23}x\left(\frac{17}{26}+\frac{9}{26}\right)\)

\(=\frac{5}{23}x1=\frac{5}{23}\)