tính tổng
A= \(\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+.......+\frac{5}{2915}\)
K
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28 tháng 10 2018
\(\frac{1}{3}+\frac{2}{15}+\frac{5}{35}+\frac{4}{63}+\frac{5}{99}\)
\(=\frac{1155}{3465}+\frac{462}{3465}+\frac{495}{3465}+\frac{220}{3465}+\frac{175}{3465}\)
\(=\frac{2507}{3465}\)
DP
30 tháng 6 2017
\(X-\left(\frac{31}{5}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\frac{9}{13}\)
\(X-\left(\frac{31}{5}+\frac{31}{3\cdot5}+\frac{31}{5\cdot7}+\frac{31}{7\cdot9}+\frac{31}{9\cdot11}+\frac{31}{11\cdot13}\right)=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\frac{10}{39}+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\frac{1984}{195}=\frac{9}{13}\)
\(\Rightarrow X=\frac{9}{13}+\frac{1984}{195}=\frac{163}{15}\)
KL
6
8 tháng 6 2017
Ta có:
\(A=\frac{5}{15}+...+\frac{5}{399}\)
\(\Rightarrow A=5.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{5}{7}\)
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!!
chúc bạn hok tốt!!
Bạn nhớ k cho mik một cái đúng nha!!
15 tháng 4 2020
Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
PM
23 tháng 4 2018
2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
=\(1-\frac{1}{15}=\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
23 tháng 4 2018
a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)
A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)
A=2(1-1/15)
A=1/2.14/15
A=7/15
9 tháng 5 2017
a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{5}{39}\)
9 tháng 5 2017
a)1/3.5+1/5.7+...+1/11.13
=1/2x(1/3-1/5+1/5-1/7+...+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
24 tháng 7 2019
\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)
\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)
\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)
\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)
b)
\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)
\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)
=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)
\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)
\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)
=> \(B=\frac{140}{69}:2=\frac{70}{69}\)
N
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
B
2 tháng 4 2019
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
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!
thank