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\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
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
\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right)x=\frac{16}{5}\)
\(\Rightarrow\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}.x=\frac{16}{5}-\frac{8}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=4\)
vậy x=4
(5/17-5/17)-(5/29-5/29)-(5/37-5/37)-(15-5)=0-0-0-10=-10