A=1/15+1/35+1/63+....+1/2915+1/3135 . A = mấy ?
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NT
1
27 tháng 3 2015
Ta có:
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
Coi \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\)
\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\right)\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}+\frac{2}{55.57}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(=\frac{1}{3}-\frac{1}{57}\)
\(=\frac{19}{57}-\frac{1}{57}=\frac{18}{570}=\frac{6}{19}\)
\(\Rightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
Vậy tổng trên bằng \(\frac{3}{19}\)
TN
4
29 tháng 9 2017
1/15 +1/35 +1/63 + 1/99 +...+1/2915 +1/3135
=1/3x5+1/5x7+1/7x9+....+1/53x55+1/55x57
=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/53-1/55+1/55-1/57
=1/3-1/57
=6/19 nhé
29 tháng 9 2017
Ta có:1/15+1/35+1/63+1/99+...+1/2915+1/3135=1/3*5+1/5*7+1/7*9+1/9*11+...+1/53*55+1/55*57
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55+1/55-1/57
=1/3-1/57=19/57-1/57=18/57
15 tháng 8 2018
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\)
\(2A=\frac{1}{3}-\frac{1}{57}\)
\(2A=\frac{6}{19}\)
\(A=\frac{3}{19}\)
15 tháng 8 2018
Ta có:
A=1/15+1/35+1/63+1/99+...+1/2015+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/45.47+1/47.49
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/45-1/47+1/47-1/49
=1/3-1/49
=49/147-3/147
=47/147
15 tháng 4 2020
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
Mik viết thế này mong bạn thông cảm nha!!
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Bạn nhớ k cho mik một cái đúng nha!!
15 tháng 4 2020
Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
1 tháng 7 2019
\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)
\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)
Còn lại em nhân giống ở trên nhé
1 tháng 7 2019
Đặt A = 1/15 + 1/35 + ... + 1/3135
A = 1/3.5 + 1/5.7 + ... + 1/55.57
2A = 2/3.5 + 2/5.7 + ... + 2/55.57
2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57
2A = 1/3 - 1/57 = 6/19
A = 3/19
XO
1 tháng 7 2019
a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)
=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)
=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)
=> \(x-\frac{11}{5}=2\)
=> \(x=2+\frac{11}{5}\)
=> \(x=\frac{21}{5}\)
CD
12
DT
22 tháng 4 2016
Ta có:
A=5/15+5/35+5/63+5/99+...+5/2915
=>A=5/3.5+5/5.7+5/7.9+5/9.11+...+5/53.55
=>A=5/2.(2/3.5+2/5.7+2/7.9+2/9.11+...+2/53.55)
=>A=5/2.(2/3-2/5+2/5-2/7+2/7-2/9+2/9-2/11+...+2/53-2/55)
=>A=5/2.(2/3-2/55)
=>A=5/2.104/165
=>A=52/33
Vậy A=52/33
OK!
S
6 tháng 4 2018
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{9.11}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(tắt 1 bước nha)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{9}-\frac{1}{11}\)
\(2A=\frac{1}{3}-\frac{1}{11}\)
\(2A=\frac{8}{33}\)
\(\Rightarrow A=\frac{4}{33}\)
Vậy A=_____________
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{53\times55}+\frac{1}{55\times57}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{55\times57}\right)\)
\(=\frac{1}{2}\times\left(\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+...+\frac{57-55}{55\times57}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{57}\right)=\frac{3}{19}\)