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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
B=\(\frac{12+\frac{4}{3}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12+\frac{12}{9}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{9}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}:\frac{4\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
=\(4:\frac{4}{5}\)
=\(5\)
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
\(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
\(A=0\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left[\frac{4.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}{1.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}:\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}\right].\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left(4:\frac{4}{5}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.5.\frac{124242423}{237373735}\)
\(B=\frac{47.5.124242423}{41.237373735}\)
\(B=\frac{29196969405}{9732323135}\)
Ủng hộ mk nha !!! ^_^
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) \(B=1\frac{6}{41}\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}.\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\frac{12}{3}.\left(\frac{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}\right).\frac{4}{5}.\left(\frac{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}\right).\frac{123}{235}\)
\(B=\frac{47.4.4.123}{41.5.235}\)
\(B=\frac{47.4.4.41.3}{41.5.47.5}\)
\(B=\frac{4.4.3}{5.5}\)
\(B=\frac{48}{25}\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}\right):\left(\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.\left[\frac{12\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right]:\left[\frac{4\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right].\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}\left(\frac{12}{3}.\frac{5}{4}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.5.\frac{123}{235}\)
\(B=\frac{47.5.123}{41.235}=\frac{47.5.41.3}{41.5.47}=3\)
B=\(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)
B=\(\frac{47}{41}.\left(\frac{12.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}:\frac{4.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right).\frac{123.1010101}{235.1010101}\)
B=\(\frac{47}{41}.\left(\frac{12}{3}:\frac{4}{5}\right).\frac{123}{235}=\frac{47}{41}.\left(\frac{12}{3}.\frac{5}{4}\right).\frac{123}{235}\)
B=\(\frac{47}{41}.\frac{15}{3}.\frac{123}{235}=\frac{47.5.3.41.3}{41.3.5.47}=3\)
Vậy B=3
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