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c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}\right):\left(\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.\left[\frac{12\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right]:\left[\frac{4\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right].\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}\left(\frac{12}{3}.\frac{5}{4}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.5.\frac{123}{235}\)
\(B=\frac{47.5.123}{41.235}=\frac{47.5.41.3}{41.5.47}=3\)
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
Chúc bn học tốt
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
\(\frac{4.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}{3.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}-\frac{\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}}{3.\left(\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}\right)}=\frac{4}{3}-\frac{1}{3}=\frac{3}{3}=1\)