a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)

\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=1-\frac{1}{25}\)

\(=\frac{24}{25}\)

đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)

\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)

\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)

\(2A=\frac{1}{9}-\frac{1}{45}\)

\(2A=\frac{4}{45}\)

\(A=\frac{4}{45}\div2\)

\(A=\frac{2}{45}\)

\(\text{Ta có: }\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)

\(=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{8}\right)-......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)

\(=1-\frac{1}{1024}\)

\(=\frac{1023}{1024}\)

nay tui con thay

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)

sửa lại đề 1 chút : 

\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{98\cdot101}\)

\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{98\cdot101}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{1}{3}\cdot\frac{99}{202}\)

\(=\frac{33}{202}\)

= 1/3.(1/2-1/5)+1/3.(1/5-1/8)+....+1/3.(1/92-1/95)+1/3.(1/95-1/98)

=1/3.(1/2-1/5+1/5-1/8+....+1/92-1/95+1/95-1/98)

=1/3.(1/2-1/98)

=1/3.24/49

=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)

1/2.5+...+1/95.98

=1/2-1/5+1/5-1/8+....1/95+1/98

=1/2-1/98

=24/49

K NHA!!!

Mk bik câu B nè!

2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99

2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99

2B = 1/3 - 1/99

2B = 32/99

=> B = 16/99 

Bạn có chắc là đúng ko vậy

bài A sai đề rồi bạn coi lại đi

B=1/3.5+1/5.7+1/7.9+...........+1/97.99

B=1.2/2.3.5+1.2/2.5.7+1.2/2.7.9+...+1.2/2.97.99

B=1/2.2/3.5+1/2.2/5.7+1/2.2/7.9+...+1/2.2/97.99

B=1/2(1/3.5+1/5.7+1/7.9+...+1/97.99)

B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)    (dùng phương pháp khử)

B=1/2(1/3-1/99)

B=1/2.32/99

B=16/99

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

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

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

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

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

\(A=\frac{16}{98}=\frac{8}{49}\)

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

\(B=2\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}\right)\)

\(B=2\left[\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\right]\)

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

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

\(B=2\left[\frac{1}{3}\cdot\frac{99}{100}\right]\)

\(B=2\cdot\frac{33}{100}\)

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

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

3A = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98

3A = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98

3A = 1/2 - 1/98

3A = 24/49

A = 24/49 : 3

A = 72/49

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

3/2B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100

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

3/2B = 1 - 1/100

3/2B = 99/100

B = 99/100 : 3/2

B = 33/50

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

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

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)

\(\Rightarrow A=\frac{24}{49}:3=\frac{8}{49}.\)

Vậy \(A=\frac{8}{49}.\)

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

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow B=\frac{99}{100}:\frac{3}{2}=\frac{33}{50}.\)

Vậy \(B=\frac{33}{50}.\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)

<=>\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{98}{1545}\)

<=>\(\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{x+3}\right)=\frac{98}{1545}\)

<=>\(\frac{1}{3}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}\)

<=>\(\frac{1}{3}-\frac{1}{x+3}=\frac{98}{515}\)

<=>\(\frac{1}{x+3}=\frac{1}{3}-\frac{98}{515}\)

<=>\(\frac{1}{x+3}=\frac{221}{1545}\)

<=> \(x=?????\)

Hình như đề sai hay sao vậy bạn?

hai dòng dưới đề ý trong ngoặc lúc đầu đâu có \(\frac{1}{3}\)

mà cảm ơn vì đã giúp mk mk sẽ tk cho bn

1 +( -2) + 3 + (-4) +...+2001 + (-2002) + 2003

= [1 +( -2)] + [3 + (-4)] +...+ [-2000+2001] + [(-2002) + 2003]

= -1 + -1 +............ + 1 + 1

= 0