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

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)

\(\Rightarrow x+2=99\)

\(\Rightarrow x=99-2\)

\(\Rightarrow x=97\)

Vậy \(x=97\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)

\(\Rightarrow x+2=99\)

\(\Rightarrow x=99-2\)

\(\Rightarrow x=97\)

Vậy x=97

=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))

=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))

=>A=\(\frac{350}{101}\)

7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))

7/2 ( 1 - 1/101 ) 

7/2 x 100/101

=350/101 

= 7 phần 1 + 7 phần 3 - 7 phần 3 + 7 phần 5 - 7 phần 5 + 7 phần 7 + ...... + 7 phần 99 - 7 pjaafn 101

= 7 - 7 phần 101 

 = tự tính kết quả nhé , mà bạn viết ra giấy rồi viết vào vở cho dễ nhìn nhé

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

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

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

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{101}-\frac{1}{103}\)

\(A=\frac{1}{3}-\frac{1}{103}\)

\(A=\frac{100}{309}\)

\(A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}+\frac{2}{101\times103}\)

\(A=1\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}+\frac{1}{101}-\frac{1}{103}\right)\)

\(A=1\times\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(A=1\times\frac{100}{309}\)

\(A=\frac{100}{309}\)

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)

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

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

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

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

lộn:

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

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

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

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

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

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

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=1-\frac{1}{100}\Rightarrowđpcm\)

Ta có :

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

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

\(=1-\frac{1}{101}< 1\)\(\left(đpcm\right)\)

=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101=1-1/101=100/101<1

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)

\(=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)

10/11

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)