Tính tổng:
F=2.5+5.8+............+101.104
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LT
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2024
\(P=2.5+5.8+8.11+...+101.104\)
\(\Rightarrow9P=2.5.9+5.8.9+8.11.9+...+101.104.9\)
\(\Rightarrow9P=2.5.9+5.8.\left(11-2\right)+8.11.\left(14-5\right)+...+101.104.\left(107-98\right)\)
\(\Rightarrow9P=2.5.9-2.5.8+5.8.11-5.8.11+8.11.14+...-98.101.104+101.104.107\)
\(\Rightarrow9P=2.5.9-2.5.8+101.104.107\)
\(\Rightarrow9P=1123938\)
\(\Rightarrow P=124882\)
DV
7 tháng 3 2018
3S=3/2.5+3/5.8+3/8.11+...+3/101.104
3S=1/2-1/5+1/5-1/8+1/8-1/11+...+1/101-1/104
3S=1/2-1/104
S=51/104:3
S=17/104
Vậy S=17/104
7 tháng 3 2018
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{101.104}\)
\(\Rightarrow3S=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{101.104}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{101.104}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{101}-\frac{1}{104}\)
\(=\frac{1}{2}-\frac{1}{104}\)
\(=\frac{51}{104}\)
\(\Rightarrow S=\frac{51}{104}:3=\frac{51}{104}.\frac{1}{3}\)
\(=\frac{7}{104}\)
VẬY \(S=\frac{7}{104}\)
NP
1
NP
20 tháng 3 2016
1938.(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938.3/3(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938/3.(3/2.5+3/5.8+3/8.11+...+3/17.20)
= 1938/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11+...+1/17-1/20)
= 1938/3.(1/2-1/20)
= 1938/3.9/20
= 2907/10
TN
2
1 tháng 7 2016
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{95}-\frac{1}{98}\)
\(=\frac{1}{2}-\frac{1}{98}\)tự làm tiếp
TN
5
1 tháng 7 2016
A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98
A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )
A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )
A = 1/3 . ( 1/2 - 1/98 )
A = 1/3 . 24/49
A = 8/49
1 tháng 7 2016
A = 1/2x5 + 1/5x8 + 1/8x11 + ... + 1/92x95 + 1/95x98
A = 1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ... + 3/92x95 + 3/95x98 )
A = 1/3 x ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/95 - 1/98 )
A = 1/3 x ( 1/2 - 1/98 )
A = 1/3 x 24/29
A = 8/49
VH
1
LT
1
15 tháng 5 2016
\(\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+...+\frac{2}{302\cdot305}\)
=\(\frac{2}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{302\cdot305}\right)\)
=\(\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
=\(\frac{2}{3}\left(\frac{1}{2}-\frac{1}{305}\right)\)
=\(\frac{2}{3}\cdot\frac{303}{610}\)
=\(\frac{101}{305}\)
LN
31 tháng 1 2019
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
T
3
DP
26 tháng 7 2017
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
MH
25 tháng 6 2020
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)