=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

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

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

Lấy số đầu + số cuối :3+1

=21​−51​+51​−71​+....+951​−981​

=12−198=21​−981​tự làm tiếp

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 tick cho tui

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

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

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

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 = \(\frac{8}{49}\)

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

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

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

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

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

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

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

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

Vậy \(A=\frac{8}{49}\)

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

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

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

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

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

Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)

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

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

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

\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)

    \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{98}\)

\(=\dfrac{24}{49}\)

\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

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

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

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

\(=\dfrac{1}{3}\cdot\dfrac{96}{196}=\dfrac{8}{49}\)

\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)

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

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

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\cdot\frac{93}{190}\)

\(=1-\frac{31}{190}\)

\(=\frac{159}{190}\)

\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)

\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\frac{93}{190}\)

\(=1-\frac{31}{190}\)

\(=\frac{159}{190}\)

\(1-\frac{1}{2.5}-\frac{1}{5.8}-..-\frac{1}{92.95}=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{92.95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\frac{93}{190}=1-\frac{31}{190}=\frac{159}{190}\)

   học tốt nha

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

= 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

công chúa ánh trăng tim cậu bỏ 1/92.95 đi đâu vậy