3.25 ÷ 0.25 + 6 × 3.25 =32,5

\(=3,25\times4+6\times3,25\)

\(=3,25\times\left(4+6\right)\)

\(=3,25\times10\)

\(=32,5\)

=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

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

\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

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

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

\(A=\frac{1}{3}\left(\frac{49}{98}-\frac{1}{98}\right)\)

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

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

Vậy A = \(\frac{8}{49}\)

Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 =  1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)

bài này có tính chất rồi mà 

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 \(A=\frac{8}{49}\)

C = 6/2.5 + 6/5.8 + 6/8.11 +...+ 6/29.32
C = 2.(3/2.5 + 3/5.8 + 3/8.11 + ... + 3/29.32)
C = 2.(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/29 - 1/32)
C = 2.(1/2 - 1/32)
C = 2.15/32
C = 15/16

Con cặc

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

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

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

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

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

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

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

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

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

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

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

Nhân cả 2 vế với 3 rồi làm như tính dãy số có quy luật ế