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19 tháng 4 2019

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

19 tháng 4 2019

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{99}{100}\)

16 tháng 4 2018

Bài 1:

a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) ta có: \(A=1+2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2019}\)

\(\Rightarrow2A-A=2^{2019}-2\)

\(\Rightarrow A=2^{2019}-2\)

Chúc bn học tốt !!!!!

16 tháng 4 2018

a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

25 tháng 3 2018

A= \(\frac{1}{1.2}\)\(\frac{1}{2.3}\)\(\frac{1}{3.4}\)\(\frac{1}{4.5}\)\(\frac{1}{5.6}\)+......+ \(\frac{1}{99.100}\)

A = \(\frac{1}{1}\)\(\frac{1}{2}\)\(\frac{1}{2}\)\(\frac{1}{3}\)+ 1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6+.....+ 1/99 - 1/ 100

A= 1/1 - 1/100

A= 99/100

k cho mik nhé

1 tháng 7 2019

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

10 tháng 4 2018

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

         \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

           \(=1-\frac{1}{100}\)

            \(=\frac{99}{100}\)

21 tháng 5 2019

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

\(B=1-\frac{1}{100}=\frac{99}{100}\)

~ Hok tốt ~

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\)  = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}\)\(1-\frac{1}{26}=\frac{25}{26}\)

~Học tốt~

27 tháng 3 2019

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{25.26}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{25}-\frac{1}{26}\)

\(=1-\frac{1}{26}=\frac{25}{26}\)

7 tháng 5 2018

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\frac{1}{100}\)

\(B=\frac{99}{100}\)

7 tháng 5 2018

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\) 

  \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

  \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

  \(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)

    \(=1-\frac{1}{100}=\frac{99}{100}\)

13 tháng 4 2018

Ta có : 

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

Chúc bạn học tốt ~ 

13 tháng 4 2018

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(\Leftrightarrow A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)

\(\Leftrightarrow A=\frac{49}{100}\)

Vậy A=\(\frac{49}{100}\)

22 tháng 6 2018

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}+x=100\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}+x=100\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}+x\right)=100\)

\(\left(1-\frac{1}{100}\right)+x=100\)

\(\frac{99}{100}+x=100\)

\(x=100-\frac{99}{100}=\frac{9901}{100}\)

22 tháng 6 2018

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}+x=100\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+....+\frac{1}{99}-\frac{1}{100}+x=100\)

\(\Rightarrow1-\frac{1}{100}+x=100\)

\(\Rightarrow\frac{99}{100}+x=100\)

\(\Rightarrow x=100-\frac{99}{100}\)

\(\Rightarrow x=\frac{1}{100}\)

~Chúc bạn hok tốt~

28 tháng 6 2020

Bài làm 

\(D=\frac{6}{3,5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)

\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)

\(D=3.\frac{20}{69}\)

\(D=\frac{20}{23}\)

Học tốt 

28 tháng 6 2020

Bài làm 

 \(D=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)

\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)

\(D=3.\frac{20}{69}\)

\(D=\frac{20}{23}\)

   \(E=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(E=10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)

\(E=10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(E=10.\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(E=10.\frac{4}{55}\)

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

     \(G=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(G=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(G=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(G=\frac{1}{1}-\frac{1}{100}\)

\(G=\frac{99}{100}\)

Nhớ k cho m nha