1phần 7 và 2
K
Khách
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18 tháng 7 2020
Nhân 2 bên với 4 được:
\(4E=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{95\cdot99}\)
\(4E=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
\(4E=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(E=\frac{\frac{32}{99}}{4}=\frac{8}{99}\)
18 tháng 7 2020
Bg
Ta có: E = \(\frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+...+\frac{1}{95\times99}\)
=> E = \(\frac{1}{4}\times\left(\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{95\times99}\right)\)
=> E = \(\frac{1}{4}\times\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)
=> E = \(\frac{1}{4}\times\left(\frac{1}{3}-\frac{1}{99}\right)\)
=> E = \(\frac{1}{4}\times\left(\frac{33}{99}-\frac{1}{99}\right)\)
=> E = \(\frac{1}{4}\times\frac{32}{99}\)
=> E = \(\frac{8}{99}\)
16 tháng 5 2021
7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0
7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)
7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)
7/48 - 193/56 : x = 0
193/56 : x = 0 + 7/48
193/56 : x = 7/48
x = 193/56 : 7/48
x = 1158/49
ND
0
MV
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ =1-\dfrac{1}{2^{99}}< 1\)
Vậy \(B< 1\)
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\dfrac{1}{2^{99}}\)
\(\rightarrow B< 1\rightarrowđpcm\)
NT
5
NM
1
17 tháng 7 2020
Bài làm:
Ta có: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{98}{99}+\frac{1}{2}.\frac{49}{100}\)
\(=\frac{49}{99}+\frac{49}{200}\)
\(=\frac{14651}{19800}\)
\(2\frac{1}{7}=\frac{15}{7}\)