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

= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)

= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

= 2.(1/1-1/100)

= 2.(99/100)

=99/50

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

\(A=\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{97\cdot100}\right)\)

\(A=\dfrac{2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}\cdot\left(1-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}\cdot\dfrac{99}{100}\)

\(A=\dfrac{33}{50}\)

\(A=\dfrac{2}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\times\dfrac{99}{100}=\dfrac{33}{50}\)

Mình mới học lớp 5 , xin lỗi nhé, mình cũng rất muốn giúp bạn nhưng ko đc.

nếu không làm được thì thôi, mong bạn đừng nhắn lời xin lỗi ạ. Không ai như bạn đâu!

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

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

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

\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{2}{3}.\frac{99}{100}\)

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

Ta có: \(A=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

Nhận xét: \(\frac{a}{x.\left(x+a\right)}=\frac{1}{x}-\frac{1}{x+a}\)

Do đó: \(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(\frac{100}{100}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\frac{99}{100}\)

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

Vậy,\(A=\frac{33}{50}\)

\(\text{Ta có: }A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)

\(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)

\(\Rightarrow\frac{3}{2}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow\frac{3}{2}A=1-\frac{1}{100}\)

\(\Rightarrow\frac{3}{2}A=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}:\frac{3}{2}\)

\(A=\frac{99}{100}.\frac{2}{3}=\frac{33}{50}\)

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

A = 2/3.3/1.4 + 2/3.3/4.7 + 2/3.3/97.100

A = 2/3( 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100 ( dùng phương pháp khử )

A = 2/3(1 - 1/100 )

A = 2/3.99/100

A = 33/50

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

=2/3(3/1.4+3/4.7+3/7.10+...+3/97.100)

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

=2/3(1-1/100)=33/50

\(S=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+......+\frac{2}{97.100}\)

\(\Rightarrow S=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\\ \Rightarrow S=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\\ \Rightarrow S=\frac{2}{3}\left(1-\frac{1}{100}\right)\\ \Rightarrow S=\frac{33}{50}\)

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

=2/3 (3/1.4+3/4.7+3/7.10+...+3/97.100)

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

=2/3 (1-1/100)=33/50

ok rồi ạ

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

tại sao ko đóng mở ngoặc phép tính rồi nhân 3 vậy?

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=\frac{297}{100}\)

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

= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)

= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

= 2.(1/1-1/100)

= 2.(99/100)

=99/50

\(\frac{2}{1.4}+\frac{2}{4.7}+....+\frac{2}{97.100}\)

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

\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\left(\frac{200}{100}-\frac{2}{100}\right)=\frac{1}{3}.\frac{198}{100}=\frac{33}{50}\)

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

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

\(=\frac{2}{3}.\frac{99}{100}\)

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