s=(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)+(1/103-1/104+1/104-1/105+1/105-1/106+1/106-1/107)

  =(1-1/103)+(1/103-1/107)

  =1           -         1/107

  =106/107

Trong trường hợp bn viết nhầm 1/106.107 chứ ko phải 1/106+107

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106.107}\)

\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}+\frac{1}{106}-\frac{1}{107}\right)\)

\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{107}\right)\)

\(S=\frac{102}{103}+\frac{4}{11021}\)

\(S=\frac{106}{107}\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106+107}\)

\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}\right)+\frac{1}{106+107}\)

\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{106}\right)+\frac{1}{106+107}\)

\(S=\frac{102}{103}+\frac{3}{10918}+\frac{11343}{106}\)

\(S=108\)

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

\(A=3.\dfrac{99}{100}\)

\(A=\dfrac{297}{100}\)

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

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

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

`#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`

\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)

\(\dfrac{34}{103}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{1}{1}-\dfrac{1}{103}\)

\(=\dfrac{102}{103}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{102}{309}=\dfrac{34}{103}\)

\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\frac{102}{103}\)

\(B=\frac{34}{103}\)

Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)

\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)

Thằng an mất dạy :))) 
#Lê_Linh

=1/1-1/4+1/4-1/7+....+1/n-1/n+3

=1-1/n+3

=n+2/n+3

Ta có : 

3/ 1.4 + 3/ 4.7 + 3/ 7.10 + ... + 3/ n( n + 1 )

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/ n - 1/ n + 3 .

= 1 - 1/ n+3 .

= n+3 - 1 / n+3 

= n+2 / n+3 .

\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n.\left(n+3\right)}\right)\)

\(3.\left(1.\frac{1}{4}+\frac{1}{4}.\frac{1}{7}+\frac{1}{7}.\frac{1}{10}+...+\frac{1}{n}.\frac{1}{n+3}\right)\)

\(3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\right)\)

\(3.\left(1-\frac{1}{n+3}\right)=3.\frac{n+2}{n+3}=\frac{6n+6}{n+3}\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}