\(A=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)

\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(A=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(A=7\left(\dfrac{3}{35}\right)\)

\(A=\dfrac{3}{5}\)

Cách giải vậy đó

A=7/3(1/10-1/11+1/11-1/12+......+1/69-1/70

A=7/3(1/10-1/70)

A=7/3.3/35

A=1/5

lê mạnh tiến đạt làm sai rồi

sao lại đưa 7 ra ngoài phải đưa 7/3 ra ngoài mà

\(C=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)

= \(7\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{69.70}\right)\)

= \(7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

= \(7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

=\(7\left(\dfrac{7}{70}-\dfrac{1}{70}\right)\)

= \(7.\dfrac{6}{70}\)

= \(\dfrac{3}{5}\)

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=2.\dfrac{3}{16}\)

\(A=\dfrac{3}{8}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{111}\)

\(B=\dfrac{12}{37}\)

\(C=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)

\(C=7\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{7}{69.70}\right)\)

\(C=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(C=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(C=7.\dfrac{3}{35}\)

\(C=\dfrac{3}{5}\)

1) A=7/10.11+7/11.12+7/12.13+...+7/69.70

  A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)

 A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)

 A= 7.(1/10-1/70)

 A=7.3/35=3/5

2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75

  B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

  B=1/25-1/75=2/75

A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70

(1/7).A=1/10.11+1/11.12+...+1/69.70

=1/10-1/11+1/11-1/12+...+1/69-1/70

=1/10-1/70=3/35

=>A=7.(3/35)

=3/5

2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75

=>(1/2).B=2/25.27+...+2.73.75

=1/25-1/27+...+1/73-1/75

=1/25-1/75

=2/75

=>B=4/75

 A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)

 A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)

 A= 7.(1/10-1/70)

 A=7.3/35=3/5

2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75

  B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

  B=1/25-1/75=2/75

`=>2B=(2)/(25.27)+(2)/(27.29)+(2)/(29.31)+....+(2)/(73.75)`

`=>2B=(1)/(25)-(1)/(27)+(1)/(27)-(1)/(29)+(1)/(29)-(1)/(31)+.....+(1)/(73)-(1)/(75)`

`=>2B=(1)/(25)-(1)/(75)`

`=>2B=(3)/(75)-(1)/(75)=(2)/(75)`

`=>B=(2)/(75):2`

`=>B=1/75`

\(B=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)

\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)

Ta có: \(B=\dfrac{1}{25\cdot27}+\dfrac{1}{27\cdot29}+\dfrac{1}{29\cdot31}+...+\dfrac{1}{73\cdot75}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{25\cdot27}+\dfrac{2}{27\cdot29}+\dfrac{2}{29\cdot31}+...+\dfrac{2}{73\cdot75}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{75}=\dfrac{1}{75}\)

Gọi \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)

là A, ta có

\(A=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)

\(\Rightarrow2.A=\dfrac{2}{25.27}+\dfrac{2}{27.29}+\dfrac{2}{29.31}+...+\dfrac{2}{73.75}\)\(\Rightarrow2.A=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2.A=\dfrac{1}{25}-\dfrac{1}{75}\)

\(\Rightarrow2.A=\dfrac{2}{75}\)

\(\Rightarrow A=\dfrac{2}{75}\div2\)

\(\Rightarrow A=\dfrac{1}{75}\)

KL: Vậy A =\(\dfrac{1}{75}\)

Tính :

a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

thanks

M=2/1.3+2/3.5+2/5.7+...+2/99.101

    =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =101/101-1/101

 M = 99/101

A=7/10.11+7/11.12+7/12+13+7/69/70 ( sai đề )

 B= 1/25.27+1/27.29+1/29.31+...+1/73.75

   =1/2.(2/25.27+2/27.29+2/29.31+...+2/73.75

   =1/2.(1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

   =1/2.(1/25-1/75)

   =1/2.75/100

   =3/8