\(\dfrac{18}{2.5}+\dfrac{18}{5.8}+....+\dfrac{18}{203.206}\)
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NV
4
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
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
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)
OY
5 tháng 9 2021
Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)
\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)
5 tháng 9 2021
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(=\dfrac{1}{2}-\dfrac{1}{98}\)
\(=\dfrac{24}{49}\)
O
26 tháng 3 2022
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{92}+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(A=\dfrac{49}{98}-\dfrac{1}{98}\)
\(A=\dfrac{48}{98}\)
\(A=\dfrac{24}{49}\)
26 tháng 3 2022
Giải thích các bước giải:
A =1/2.5 + 1/5.8 + 1/8.11 + … +1/92.95 + 1/95.98
=1/3 . (1/2-1/5+1/5-1/8+1/8-1/11+…+1/92-1/95+1/95-1/98)
=1/3 . (1/2 – 1/98 )
=1/3 . 24/49
=8/49`
vậy `A=8/49`
1 tháng 4 2023
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
24 tháng 7 2021
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
PH
1
HT
17 tháng 7 2015
\(\frac{18}{2.5}+\frac{18}{5.8}+....+\frac{18}{103.106}\)
=\(6\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{103.106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{103}-\frac{1}{106}\right)\)
=\(6\left(\frac{1}{2}-\frac{1}{106}\right)\)
=\(6.\frac{26}{53}\)
=\(\frac{156}{53}\)
\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)
\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)
\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(A=6.\dfrac{51}{103}\)
\(A=\dfrac{306}{103}\)
18/(2.5) + 18/(5.8) + .... + 18/(203.206)
= 18.[1/(2.5) + 1/(5.8) + .... + 1/(203.206)]
= 18.(1/2 - 1/5 + 1/5 - 1/8 + .... + 1/203 - 1/206)
=18.(1/2 - 1/206)
=18.(51/103)
=918/103