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10 tháng 5 2019

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{95.98}\)

\(A=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{2}{3}.\frac{24}{49}\)

\(A=\frac{16}{49}\)

11 tháng 5 2019

\(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{95.98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{48}{98}=\frac{24}{49}\)

\(\Leftrightarrow A=\frac{24}{49}\div\frac{3}{2}\)

\(\Leftrightarrow A=\frac{48}{147}\)

3 tháng 5 2022

nhân 3 vào cả hai vế 

3 tháng 5 2022

xong tự tính

2 tháng 5 2018

3A = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\)

3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)

3A=\(\frac{1}{2}-\frac{1}{98}\)

3A=\(\frac{98}{196}-\frac{2}{196}\)=\(\frac{96}{196}=\frac{24}{49}\)

A=\(\frac{24}{49}:3=\frac{24}{49}.\frac{1}{3}=\frac{8}{49}\)

Vậy A = \(\frac{8}{49}\)

2 tháng 5 2018

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)

\(\Rightarrow3A=3\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\right)\)

\(\Rightarrow3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}\)

\(\Rightarrow3A=\frac{24}{49}\)

\(\Rightarrow A=\frac{24}{49}:3\)

\(\Rightarrow A=\frac{8}{49}\)

Vậy \(A=\frac{8}{49}\)

18 tháng 9 2015

bài này có tính chất rồi mà 

21 tháng 5 2018

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{95\cdot98}\)

\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{95\cdot98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{48}{98}\)

\(A=\frac{16}{98}=\frac{8}{49}\)

\(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(B=2\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}\right)\)

\(B=2\left[\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\right]\)

\(B=2\left[\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\right]\)

\(B=2\left[\frac{1}{3}\left(1-\frac{1}{100}\right)\right]\)

\(B=2\left[\frac{1}{3}\cdot\frac{99}{100}\right]\)

\(B=2\cdot\frac{33}{100}\)

\(B=\frac{33}{50}\)

21 tháng 5 2018

A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

3A = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98

3A = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98

3A = 1/2 - 1/98

3A = 24/49

A = 24/49 : 3

A = 72/49

B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/97.100

3/2B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100

3/2B = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100

3/2B = 1 - 1/100

3/2B = 99/100

B = 99/100 : 3/2

B = 33/50

Áp dụng ct : 1/n.(n+1) = 1/n - 1/n+1

Ta có : A = 1/2.5 + 1/5.8 + ...+1/95.98

           A = 1/2 - 1/5 + 1/5 - 1/8 +...+ 1/95 - 1/98

           A = 1/2 - 1/98 

           A = 24/49

k mk nha bn

24 tháng 4 2016

= 1/3 . (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 + 1/11 - ... + 1/92 - 1/95)

= 1/3 . (1/2 - 1/95)

= 1/3 . 93/190

= 31/190

tớ chắc nha nguten duc huy

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`

3 tháng 5 2018

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)

3 tháng 5 2018

\(=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=3.\frac{24}{49}\)

\(=\frac{72}{49}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{95}-\frac{1}{98}\)

\(=\frac{1}{2}-\frac{1}{98}\)tự làm tiếp

11 tháng 5 2022

Lấy số đầu + số cuối :3+1