= 5-7+7-9+9-11+...+25-27

= 5-27

= -22

5.7+7.9+9.11+...+25.27

=5.(5+2)+7.(7+2)+9.(9+2)+...+25.(25+2)

=5²+2.5+7²+2.7+9²+2.9+...+25²+2.25

=(5²+7²+9²+...+25²)+(2.5+2.7+2.9+...+2.25)

=2915+2.(5+7+9+...+25)

=2915+2.165

=2915+330

=3245

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

\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)

\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)

\(=4.\frac{2}{11}\)

\(=\frac{8}{11}\)

84/305 nha bạn

mk nha có cần giải ra ko

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

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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

N=3/2.(1/5.7+1/7.9+1/9.11+....+1/197.199)

N=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+....+1/197-1/199)

N=3/2.(1/5-1/199)

N=3/2.194/995

N=291/995

Có N = 3 . (1/5 .  7 + 1/7 . 9 + .... + 1/197 . 199)

    2N =  3 . (2/5 . 7 + 2/7 . 9 + ... + 2/197 . 199)

    2N = 3 . (1/5 - 1/7 + 1/7 - 1/9 + ... + 1/197 - 1/199)

    2N = 3 . (1/5 - 1/199)

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

      N = (3 . 194/995) : 2

Bạn tự tính nốt nhé chúc bạn học tốt