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3 tháng 5 2018

x-1/2*(1/1-1/3)-(1/3-1/5)-...-1/97-1/99=5/6

x-1/2*(1-1/99)=5/6

x-1/2*98/99=5/6

x-49/59=5/6

x=5/6+49/59=263/198

18 tháng 8 2021

a. \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3-1}{3}=\dfrac{2}{3}\)\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5-3}{15}=\dfrac{2}{15}\)

b. Ta có \(VP=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\) mà \(VP=\dfrac{2}{3}\) \(\Rightarrow VT=VP\)

Ta có \(VP=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\) mà \(VP=\dfrac{2}{3.5}=\dfrac{2}{15}\) \(\Rightarrow VT=VP\)

c. \(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)

\(=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}+\dfrac{1}{99.101}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\left(1-\dfrac{1}{101}\right)\) \(=\dfrac{200}{101}\)

a: \(\dfrac{1}{1}-\dfrac{1}{3}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)

b: \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)

c: Ta có: \(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

11 tháng 5 2019

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

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

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

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

~ Hok tốt ~

\(\)

11 tháng 5 2019

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

27 tháng 6 2015

 

\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)

\(\frac{99}{20}-2x=\frac{49}{99}\)

\(2x=\frac{99}{20}-\frac{49}{99}\)

\(2x=\frac{8821}{1980}\)

\(x=\frac{8821}{1980}:2\)

\(x=\frac{8821}{3960}\)

11 tháng 5 2018

\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}.\frac{32}{99}\)

\(\Rightarrow M=\frac{16}{99}\)

11 tháng 5 2018

\(M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5.}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{16}{99}\)

15 tháng 8 2016
  • \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)

           \(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

29 tháng 3 2018

   1/1x3 + 1/3x5 + 1/5x7 + ...............................+ 1/97x99

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

=1-1/99

=98/99

19 tháng 3 2023

\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)

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

\(B=\dfrac{1}{1}-\dfrac{1}{99}\)

\(B=\dfrac{99}{99}-\dfrac{1}{99}\)

\(B=\dfrac{98}{99}\)

#YVA

22 tháng 3 2023

B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)

B=\(\dfrac{98}{99}:2\)

B=\(\dfrac{49}{99}\)

30 tháng 3 2023

  A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)

2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)

2A= 1-\(\dfrac{1}{99}\)

2A= \(\dfrac{98}{99}\)

  A= \(\dfrac{98}{99}\) : 2

A=\(\dfrac{49}{99}\)