\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

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

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99 
           =2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
           =2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
           =1+0+0+0+...+0+1/99
           =1+1/99
           =100/99
Mà 100/99>1.Suy ra S>1
   Vậy S>1
            

S=1-1/3 + 1/3 - 1/5 + ... + 1/97 - 1/99

=1 - 1/99 => S<1

Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)

Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)

=> A = 98B

các bạn có  về sweet home

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)\(=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{99-97}{97\cdot99}\)\(=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+...+\frac{99}{97\cdot99}-\frac{97}{97\cdot99}\)\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)\(=\frac{1}{3}-\frac{1}{99}\)\(=\frac{32}{99}>\frac{8}{25}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{32}{99}\)

Nhận thấy : \(\frac{32}{99}>\frac{8}{25}\left(32>8;99>25\right)\)

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

 vipboyss5: \(\frac{32}{99}\)chứ ko phải 33

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

=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99

=1/3-1/99

=33/99-1/99

=32/99

M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)

M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99

M=1/3-1/99

M=32/99

\(\frac{32}{99}\)

\(S=\dfrac{2^2}{3x5}+\dfrac{2^2}{5x7}+\dfrac{2^2}{7x9}+...+\dfrac{2^2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

S=\(\dfrac{64}{99}\)