DT
2
K
Khách
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MB
3
19 tháng 4 2018
\(\frac{3}{2.5}\)+ \(\frac{3}{5.8}\)+ ...... + \(\frac{3}{92.95}\)= 3 . ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ .... + \(\frac{1}{92.95}\))
= 3 . \(\frac{1}{3}\). ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ ..... + \(\frac{1}{92.95}\))
= 3. \(\frac{1}{3}\). ( \(\frac{1}{2}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{8}\)+ ....... + \(\frac{1}{92}\)- \(\frac{1}{95}\))
= 1 .( \(\frac{1}{2}\)- \(\frac{1}{95}\)) = \(\frac{93}{190}\)
Thấy hay thì cho mình một k nhé!!!
KT
19 tháng 4 2018
3/ 2.5 + 3/ 5.8 + 3/ 8.11+ ...+ 3/ 92.95
=1/2-1/5+1/5-1/8+1/8-1/11+........+1/92-1/95
=1/2-1/95
=31/60
TN
2
1 tháng 7 2016
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{95}-\frac{1}{98}\)
\(=\frac{1}{2}-\frac{1}{98}\)tự làm tiếp
SK
1
QD
26 tháng 4 2017
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
TN
5
1 tháng 7 2016
A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98
A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )
A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )
A = 1/3 . ( 1/2 - 1/98 )
A = 1/3 . 24/49
A = 8/49
1 tháng 7 2016
A = 1/2x5 + 1/5x8 + 1/8x11 + ... + 1/92x95 + 1/95x98
A = 1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ... + 3/92x95 + 3/95x98 )
A = 1/3 x ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/95 - 1/98 )
A = 1/3 x ( 1/2 - 1/98 )
A = 1/3 x 24/29
A = 8/49
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 = \(\frac{8}{49}\)
WH
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 \(A=\frac{8}{49}\)
14 tháng 4 2023
=1/3(3/2*5+3/5*8+...+3/95*98)
=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)
=1/3*48/98
=1/3*24/49
=8/49
24 tháng 4 2016
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
DH
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
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`
\(A=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}+\dfrac{3}{95.98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}.\dfrac{24}{49}=\dfrac{8}{49}\)
\(A=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+..+\dfrac{3}{92.95}+\dfrac{3}{95.98}\right)\)
\(A=\dfrac{3}{2.5.3}+\dfrac{3}{5.8.3}+\dfrac{3}{8.11.3}+..+\dfrac{3}{92.95.3}+\dfrac{3}{95.98.3}\)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+..+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+..+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}=\dfrac{49}{98}-\dfrac{1}{98}=\dfrac{48}{98}=\dfrac{24}{49}\)