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BN
12 tháng 2 2017
\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(=\frac{5-3}{3.5}+\frac{8-5}{3}+\frac{19-8}{8.19}+\frac{32-29}{19.32}+\frac{57-32}{32.57}+\frac{87-57}{57.87}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(=\frac{1}{3}-\frac{1}{87}=\frac{28}{87}\)
DV
2
MT
19 tháng 10 2015
\(A=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(=\frac{1}{3}-\frac{1}{87}=\frac{29}{87}-\frac{1}{87}=\frac{28}{87}\)
NT
1
LM
1 tháng 3 2017
\(C=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)\(C=\left(\dfrac{5-3}{3\cdot5}\right)+\left(\dfrac{8-5}{5\cdot8}\right)+\left(\dfrac{19-8}{8\cdot19}\right)+\left(\dfrac{32-19}{19\cdot32}\right)+\left(\dfrac{57-32}{32\cdot57}\right)+\left(\dfrac{87-57}{57\cdot87}\right)\)\(C=\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{19}\right)+\left(\dfrac{1}{19}-\dfrac{1}{32}\right)+\left(\dfrac{1}{32}-\dfrac{1}{57}\right)+\left(\dfrac{1}{57}+\dfrac{1}{87}\right)\)\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\)\(C=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
Y
3 tháng 3 2019
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )
\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)
\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)
Rồi làm tương tự cân b nha!
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)
\(+\dfrac{1}{57}-\dfrac{1}{87}\)
\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
VB
1
20 tháng 3 2019
\(S=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.85}\)
\(S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(S=\frac{1}{3}-\frac{1}{87}\)
\(S=\frac{28}{87}\)
TQ
3
20 tháng 3 2019
Ai chs opoke đại chiên lh mik nha! Đỏi lấy nick olm hoặc cho mik
7 tháng 4 2021
Ta có: \(\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{16}{122\cdot138}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{122}-\dfrac{1}{138}\)
\(=\dfrac{1}{3}-\dfrac{1}{138}\)
\(=\dfrac{46}{138}-\dfrac{1}{138}\)
\(=\dfrac{45}{138}=\dfrac{15}{46}\)
7 tháng 4 2021
Giải:
2/3.5+3/5.8+...+16/122.138
=1/3-1/5+1/5-1/8+...+1/122-1/138
=1/3-1/138
=45/138=15/46
Chúc bạn học tốt!
=\(\left(\frac{5-3}{3.5}\right)+\left(\frac{8-5}{5.8}\right)+\left(\frac{19-8}{8.19}\right)+\left(\frac{32-19}{19.32}\right)+\left(\frac{57-32}{32.57}\right)+\left(\frac{87-57}{57.87}\right)\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
=\(\frac{1}{3}-\frac{1}{87}\)
=\(\frac{28}{87}\)
Ai thấy đúng thì k cho mk nhé!