Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)

\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)

\(=\frac{1}{3}-\frac{1}{47.49}\)

\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)

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

    m = 1/3-1/99=32/99

   Sorry chị em ko làm đc câu b vì em mới học lớp 4

   k em ha

a) \(M=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{97\times99}\)

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

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

\(\Rightarrow M=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

b)  \(N=\frac{3}{5\times7}+\frac{3}{7\times9}+\frac{3}{9\times11}+...+\frac{3}{197\times199}\)

\(\Rightarrow N=3\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\)

\(\Rightarrow N=3\times\left[2\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\right]\)

\(\Rightarrow N=3\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{197\times199}\right)\)

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

\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{199}\right)\)

\(\Rightarrow N=3\times\frac{194}{995}=\frac{582}{995}\)

----Chúc em học giỏi !----

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

M=1/3-1/99

M= 33/99-1/99

M=32/99

b

N=3/2( 2/5x7+2/7x9+2/9x11+...+ 2/ 197x199)

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

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

N=3/2x194/995

N=291/995

chắc chắn đúng 100% nha bạn

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

lộn:

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

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

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

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

\(C=\frac{1}{3}-\frac{1}{97}\)

\(C=\frac{94}{291}\)

a=1/3x5+1/5x7+...+1/2003x2005

a=1x2/3x5x2+1x2/5x7x2+...+1x2/2003x2005x2

a=1/2(2/3x5+2/5x7+...+2/2003x2005)

a=1/2x(1/3-1/5+1/5-1/7+...+1/2003-1/2005)

a=1/2x(1/3-1/2005)

a=1/2x2002/6015

a=1001/6015

A = 1/3.5 + 1/5.7 + 1/7.9 + .... + 1/2003.2005 

2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + .... + 1/2003 - 1/2005

2A = 1/3 - 1/2005 = 2002/6015 

=>A = 1001/6015

\(\frac{1}{2}A=\)\(2\times\left(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{2003\times2005}\right)\)

\(\Leftrightarrow\frac{1}{2}A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{2003\times2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{2002}{6015}\)

\(\Leftrightarrow A=\frac{1001}{6015}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{48}\)

=> 2x + 3 = 48

=> 2x = 48 - 3

=> 2x = 45

=> x = 45/2

Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(\Leftrightarrow M=\frac{2}{3.\left(3+2\right)}+\frac{2}{5.\left(5+2\right)}+...+\frac{2}{97\left(97+2\right)}\)

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

\(\Leftrightarrow M=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

( Dòng thứ 2 mik làm để bạn hiểu mik đã áp dụng công thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) nên bạn ghi hay ko cx được)

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

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

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

\(M=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

gọi biểu thức đó là A

\(A=\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{2008}{6033}\right)\)

\(A=\frac{1004}{6033}\)

mink nghĩ vậy bạn ạ

C.mơn bạn nha ! ^_^

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(\Rightarrow2x=90\)

\(\Rightarrow x=45\)