Khoảng cách có rồi thì bạn áp dụng công thức : \(\frac{a}{m.n}=\frac{1}{m}-\frac{1}{n}\)(với n-m=a) là làm được

S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{93.95}+\frac{3}{95.98}+\frac{4}{98.102}+\frac{5}{102.17}+\frac{2012}{107..2119}\)

S=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+\frac{1}{102}-\frac{1}{107}+\frac{1}{107}-\frac{1}{2119}\)

S=\(\frac{1}{5}-\frac{1}{2119}\)

S=\(\frac{2114}{10595}\)

\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+..............+\frac{2}{93\cdot95}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+............+\frac{1}{93}-\frac{1}{95}\)

\(=\frac{1}{5}-\frac{1}{95}\)

S = 2/5.7+ 2/7.9+...+2/93.95

=1/5-1/7+1/7-1/9+1/11+...+1/93+1/95

=1/5-1/95

=19/95-1/95

=18/95

Bg

Ta có: S = \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

=> S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

=> S = \(\frac{1}{5}-\frac{1}{95}\)

=> S = \(\frac{19}{95}-\frac{1}{95}\)

=> S = \(\frac{18}{95}\)

Vậy S = \(\frac{18}{95}\)

\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{93\cdot95}\)

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(S=\left(\frac{1}{5}-\frac{1}{95}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{93}+\frac{1}{93}\right)\)

\(S=\left(\frac{1}{5}-\frac{1}{95}\right)\)

\(S=\frac{19}{95}-\frac{1}{95}\)

\(S=\frac{18}{95}\)

                     Bài làm :

Ta có:

 \(S=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

\(S=\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+...+\frac{95-93}{93.95}\)

 \(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

 \(S=\frac{1}{5}-\frac{1}{95}\)

\(S=\frac{18}{95}\)

 \(\text{Vậy S =}\frac{18}{95}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

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

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(\dfrac{1}{3}-\dfrac{1}{101}\)

= \(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

1)           P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15

              P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)

              P=1/3-1/15= 4/15

2) a/ 0,2:1+3/5+80%

= 2/10:8/5+8/10

= 2/10.5/8+8/10

= 1/8 + 4/5 = 5/40 + 32/40 = 37/40

    b/ 0,5:5/4-2+1/5

= 5/10:5/4-11/5

= 5/10.4/5-11/5

=2/5-11/5 = -9/5