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H24
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LA
21 tháng 8 2016 lúc 10:13

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)

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LA
21 tháng 8 2016 lúc 10:18

\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)

\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)

\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)

\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)

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LL
21 tháng 8 2016 lúc 20:17

= 2(2/2.4 + 2/4.6 +.....+ 2/2008.2016)

= 2(1/2 - 1/4 + 1/4 - 1/6 +....+ 1/2008 - 1/2016)

= 2(1/2 - 1/2016)

=2 . 1007/2016

=1007/1008

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PQ
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EC
4 tháng 8 2016 lúc 13:12

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{2014.2016}\)

\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)

\(=2.\frac{1007}{2016}=\frac{1007}{1008}\)

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LP
4 tháng 8 2016 lúc 13:13

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}\)

\(A=\frac{4}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)

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SG
4 tháng 8 2016 lúc 13:14

A = 4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2014.2016

A = 2.(2/2.4 + 2/4.6 + 2/6.8 + ... + 2/2014.2016)

A = 2.(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2014 - 1/2016)

A = 2.(1/2 - 1/2016)

A = 2.1/2 - 2.1/2016

A = 1 - 1/1008

A = 1007/1008

B = 1/18 + 1/54 + 1/108 + ... + 1/990

B = 1/9.(1/2 + 1/6 + 1/12 + ... + 1/110)

B = 1/9.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/10.11)

B = 1/9.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/10 - 1/11)

B = 1/9.(1 - 1/11)

B = 1/9.10/11

B = 10/99

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OE
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TN
10 tháng 5 2016 lúc 22:03

\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(K=2\times\frac{502}{1005}\)

\(K=\frac{1004}{1005}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)

\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(3F=\frac{1}{3}-\frac{1}{33}\)

\(F=\frac{10}{33}:3\)

\(F=\frac{10}{99}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}\)

\(I=\frac{2009}{2010}\)

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H24
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NH
14 tháng 8 2016 lúc 10:00

dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

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CT
14 tháng 8 2016 lúc 10:02

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A=1-\frac{1}{2010}\)

\(A=\frac{2009}{2010}\)

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PD
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H24
22 tháng 4 2019 lúc 21:11

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

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H24
22 tháng 4 2019 lúc 21:13

b,\(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

Vậy

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H24
22 tháng 4 2019 lúc 21:15

\(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\-\frac{1}{2}x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}}\)\

Vậy

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TM
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NT
2 tháng 7 2022 lúc 13:03

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)

c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

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CY
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NT
14 tháng 5 2022 lúc 8:20

a: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2-10+3}{15}=\dfrac{-5}{15}=\dfrac{-1}{3}\)

b: \(=\left(6+\dfrac{1}{8}-\dfrac{1}{2}\right)\cdot4=\dfrac{48+1-4}{8}\cdot4=\dfrac{45}{2}\)

c: \(=\dfrac{1}{4}\cdot4-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

d: \(F=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2008\cdot2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{1004}{2010}=\dfrac{1004}{1005}\)

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HA
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TL
4 tháng 3 2020 lúc 12:30

a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)

b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)

\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)

\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)

\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)

\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)

\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)

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 Khách vãng lai đã xóa
TL
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H24
10 tháng 7 2018 lúc 20:20

a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}\left(\frac{2}{75}\right)\)

\(=\frac{1}{75}\)

b, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1004}{2010}\right)\)

\(=2\left(\frac{502}{1005}\right)\)

\(=\frac{1004}{1005}\)

Tk hộ =v

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DH
10 tháng 7 2018 lúc 20:27

\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)

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KS
10 tháng 7 2018 lúc 20:29

\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{3}{75}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

Câu dưới đặt 2 ra ngoài rồi làm bình thường.

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