\(\frac{1}{10x11}\)+\(\frac{1}{11x12}\)+\(\frac{1}{12x13}\)+................................+\(\frac{1}{49}\)x\(\frac{1}{50}\)
giúp mình nha
K
Khách
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NQ
1
TN
8
15 tháng 6 2016
1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
15 tháng 6 2016
1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
8 tháng 6 2016
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
8 tháng 6 2016
= 7/10-7/11+7/11-7/12+7/12-7/13+...+7/69-7/70
=7/10-7/70
=42/70
k mk nha
M
1 tháng 5 2018
\(\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{98x99}\)= \(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+.....+\frac{1}{98}-\frac{1}{99}\)
= \(\frac{1}{11}-\frac{1}{99}\)
= \(\frac{8}{99}\)
Học tốt !
1 tháng 5 2018
\(\frac{1}{11\times12}+\frac{1}{12\times13}+\frac{1}{13\times14}+...+\frac{1}{97\times98}+\frac{1}{98\times99}\)
\(=\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{11}-\frac{1}{99}\)
\(=\frac{8}{99}\)
DP
26 tháng 6 2017
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
ML
26 tháng 6 2017
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
14 tháng 4 2019
ta có : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(\Rightarrow\)\(B=A\)
TP
1
HG
2 tháng 5 2020
A = 3/10x11 - 3/11x12 - .........- 3/99x100
A = 3 x ( 1/10x11 - 1/11x12 - ..... - 1/99x100 )
A = 3 x ( 11 - 10 / 10 x 11 - 12 - 11 / 11 x 12 - ...... - 100 - 99 /99 x 100 )
A = 3 x ( 11/10x11 - 10/10x11 + 12/11x12 - 11/11x12 + ......+ 100/99x100 - 99/99x100)
A = 3 x ( 1/10 - 1/11 + 1/11 - 1/12 + ......+ 1/99 - 1/100 )
A = 3 x ( 1/10 - 1/100)
A = 3 x 9/100
A = 27/100
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\(\frac{1}{10x11}+\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{49x50}\)
...
Sửa đề: \(\frac{1}{10x11}+\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{49x50}\)
Ta có: \(\frac{1}{10x11}+\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{49x50}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)