\(B=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+........+\frac{1}{27.30}\right)\)  

\(B=3.\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......-\frac{1}{27}+\frac{1}{27}-\frac{1}{30}\right)\) 

\(B=1.\left(\frac{1}{1}-\frac{1}{30}\right)\) 

\(B=\frac{29}{30}\)

B =\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

B = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{27}-\frac{1}{30}\)

B =\(\frac{1}{1}-\frac{1}{30}\)

B =\(\frac{29}{30}\)

Ta có:

\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\)

\(=1-\frac{1}{30}=\frac{29}{30}\)

Vậy \(B=\frac{29}{30}\)

a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)

=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))

= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))

=9(1-\(\frac{1}{100}\))

A=\(\frac{891}{100}\)

b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))

=1-\(\frac{1}{30}\)

B=\(\frac{29}{30}\)

a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)

\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9\left(1-\dfrac{1}{100}\right)\)

\(=9.\dfrac{99}{100}\)

\(=\dfrac{891}{100}\)

b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)

\(=1-\dfrac{1}{30}\)

\(=\dfrac{29}{30}\)

Đặt : \(A=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{27\cdot30}\)

\(A=\frac{1}{3}\left(\frac{5}{1}-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+...+\frac{5}{27}-\frac{5}{30}\right)\)

\(A=\frac{1}{3}\left(5-\frac{5}{30}\right)\)

\(A=\frac{1}{3}\cdot\frac{29}{6}\)

\(A=\frac{29}{18}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+....+\frac{5}{27.30}\)

\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{30-27}{27.30}\)

\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\cdot\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}\cdot\frac{29}{30}=\frac{29}{18}\)

`#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 = 1/1 -1/4 +1/4 - 1/7 +1/7 ........+1/40 - 1/43 

A = 1/1 - 1/43 

A = 42/43

A=1 - 1/4 + 1/4 - 1/7 + .... + 1/40 - 1/43

  = 1 - 1/43 

  = 42/43

=42/43 nha bn

\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)

\(A=1-\dfrac{1}{67}\) < 1

=> A<1

Ta có:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)

\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)

\(=3.\left(1-\dfrac{1}{67}\right)\)

\(=3.\dfrac{66}{67}\)

\(=\dfrac{198}{67}\)

Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)

Vậy \(A>1\)

sửa bài:

...  \(=1-\dfrac{1}{67}\)

\(=\dfrac{66}{67}\)

Vì \(\dfrac{66}{67}\) có tử nhỏ hơn mẫu nên \(\dfrac{66}{67}< 1\)

Vậy \(A< 1\)

`3/1.4+3/4.7+3/7.10+...+3/94.97`

`=1/1-1/4+1/4-1/7+1/7-1/10+...+1/94-1/97`

`=1-1/97`

`=96/97`

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)

Sửa lại đề :  \(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.14}+\frac{3}{14.17}\)

\(A=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.14}+\frac{3}{14.17}\right)\)

\(A=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

\(A=\frac{3}{3}\left(1-\frac{1}{17}\right)\)

\(A=\frac{16}{17}\)

P/S : Ở chỗ 3/11.14 có lẽ bạn ghi sai đề , mình nghĩ là 3/10.14 mới đúng

Sửa lại đề :  A=31.4+34.7+37.10+310.14+314.17A=31.4+34.7+37.10+310.14+314.17

A=3.(11.4+14.7+17.10+110.14+314.17)A=3.(11.4+14.7+17.10+110.14+314.17)

A=3.13(1−14+14−17+17−110+110−114+114−117)A=3.13(1−14+14−17+17−110+110−114+114−117)

A=33(1−117)A=33(1−117)

A=1617A=1617

P/S : Ở chỗ 3/11.14 có lẽ bạn ghi sai đề , mình nghĩ là 3/10.14 mới đúng

Ngu ms đi hỏi câu này

ÓC CHÓ

\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)

\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)

\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)

\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(=3\cdot\left(1-\dfrac{1}{16}\right)\)

\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)

\(=3\cdot\dfrac{15}{16}\)

\(=\dfrac{45}{16}\)

ngu :3

vcl