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NT
2
30 tháng 3 2015
=6.3/2.5 +6.3/5.8+...+6.3/203.206
=6(3/2.5+3/5.8+...+3/203.206)
=6(1/2-1/5+1/5-1/8+...+1/203-1/206)
=6[(1/2-1/206)+(1/5-1/5)+(1/8-1/8)+...+(1/203-1/203)]
=6(1/2-1/206)=6(103/206-1/206)=6. 102/206=6. 51/103=306/103
30 tháng 3 2015
A=6.( 3/2.5+3/5.8+...+3/203.206)
=6.(1/2-1/5+1/3-1/8+...+1/202-1/206)
=6.(1/2-1/206)=306/103
TP
2
2 tháng 3 2018
\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)
\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)
\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(A=6.\dfrac{51}{103}\)
\(A=\dfrac{306}{103}\)
30 tháng 3 2019
a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}\)
\(=\frac{128}{99}\)
KN
30 tháng 3 2019
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)
HM
0
NM
2
16 tháng 3 2016
1/2.5+1/5.8+1/8.11+...+1/152.155
=1/3(3/2.5+3/5.8+3/8.11+...+3/152.155
=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/152-1/155)
=1/3(1/2-1/155)
=1/3(155/310-2/310)
=1/3.153/310=51/310
Kết quả:51/310
SK
1
QD
26 tháng 4 2017
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
VN
Võ Ngọc Phương
VIP
25 tháng 8 2023
Sửa đề:
\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)
\(A=4.\dfrac{33}{68}\)
\(A=\dfrac{33}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
\(\frac{18}{2.5}+\frac{18}{5.8}+....+\frac{18}{103.106}\)
=\(6\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{103.106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{103}-\frac{1}{106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{106}\right)\)
=\(6.\frac{26}{53}\)
=\(\frac{156}{53}\)