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DD
Đoàn Đức Hà
Giáo viên
8 tháng 7 2021
\(A=\frac{2}{3\times7}+\frac{2}{7\times11}+\frac{2}{11\times15}+...+\frac{2}{99\times103}\)
\(2\times A=\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{99\times103}\)
\(=\frac{7-3}{3\times7}+\frac{11-7}{7\times11}+\frac{15-11}{11\times15}+...+\frac{103-99}{99\times103}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow A=\frac{50}{309}\)
PT
c) C=2 x5+5x8+8x11+...+23x26+26x29
d) D=3x7+7x11+11x15+...+43x47+47x51
giúp mik với thank các bn nhiều
0
23 tháng 4 2021
Ta có: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{23\cdot27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}=\dfrac{8}{27}\)
16 tháng 8 2021
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(=\frac{1}{3}-\frac{1}{104}=\frac{104}{312}-\frac{3}{312}=\frac{101}{312}\)
22 tháng 5 2018
a)\(\left(\frac{25}{49}+\frac{17}{39}+\frac{22}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\left(\frac{17}{39}+\frac{22}{39}\right)\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{39}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+1\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\frac{50}{49}\times\frac{41}{25}\)
\(=\frac{2050}{1225}\)
TA
1
23 tháng 2 2023
SỬa đề: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\)
=1/3-1/27
=8/27
DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
NH
4
3 tháng 8 2016
\(A=\frac{4}{3X7}+\frac{4}{7X11}+\frac{4}{11X15}+...+\frac{4}{100X104}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(=\frac{1}{3}-\frac{1}{104}\)
\(=\frac{101}{312}\)
Chúc bạn học giỏi nha!!!
K cho mik với nhé nguyen huu thuong 2005
3 tháng 8 2016
\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)
\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(A=\frac{1}{3}-\frac{1}{104}=\frac{104}{312}-\frac{3}{312}=\frac{101}{312}\)
PN
2
2 tháng 4 2022
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{31.35}+\frac{1}{35.39}\) ( . là nhân nhé )
\(=\frac{1}{4}.\left[4.\left(\frac{1}{3.7}+\frac{1}{4.11}+\frac{1}{11.15}+...+\frac{1}{31.35}+\frac{1}{35.39}\right)\right]\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{31.35}+\frac{4}{35.39}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{39}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{1}{4}.\left(\frac{13-1}{39}\right)=\frac{1}{4}.\frac{12}{39}=\frac{1}{4}.\frac{4}{13}=\frac{1}{13}\)
