Cho A = \(\frac{1}{17}+\frac{1}{19}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}+\frac{1}{33}+\frac{1}{35}+\frac{1}{37}+\frac{1}{39}\)
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21 tháng 1 2017
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
21 tháng 1 2017
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
VA
16 tháng 7 2016
\(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
\(A=0\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left[\frac{4.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}{1.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}:\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}\right].\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left(4:\frac{4}{5}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.5.\frac{124242423}{237373735}\)
\(B=\frac{47.5.124242423}{41.237373735}\)
\(B=\frac{29196969405}{9732323135}\)
Ủng hộ mk nha !!! ^_^
16 tháng 7 2016
a) \(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{10101.63.37-10101.37.63}{1+2+3+...+2006}\)
\(A=\frac{0}{1+2+3+...+2006}\)
\(A=0\)
b) \(B=1\frac{6}{41}\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}.\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\frac{12}{3}.\left(\frac{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}\right).\frac{4}{5}.\left(\frac{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}\right).\frac{123}{235}\)
\(B=\frac{47.4.4.123}{41.5.235}\)
\(B=\frac{47.4.4.41.3}{41.5.47.5}\)
\(B=\frac{4.4.3}{5.5}\)
\(B=\frac{48}{25}\)
Bổ sung . CMR A > 0,3