b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Giải:

A=5/9+2/15-6/9

   =(5/9-6/9)+2/15

   = -1/9 + 2/15

   = 1/45

B=2/7-3/8+4/7+1/7-5/8+5/15

   = (2/7+4/7+1/7) + (-3/8-5/8) +1/3

   = 1+ (-1) +1/3

   =1/3

C=3/5+1/15+1/57+1/3-2/9-3/4-1/36

   =9/15+1/15+1/57+19/57-8/36-27/36-1/36

   =(9/15+1/15)+(1/57+19/57)+(-8/36-27/36-1/36)

   =2/3+20/57+(-1)

   =58/57+(-1)

   =1/57

D=1/1.2+1/2.3+1/3.4+...+1/99.100

   =1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

   =1/1-1/100

   =99/100

Câu E mình ko biết làm nhé!

c) 

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)

   \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

   \(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)

   \(=\frac{1}{2}.\frac{20}{21}\)

   \(=\frac{10}{21}\)

\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)

a)

 \(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{49.50}\)

    \(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

    \(=\frac{1}{3}-\frac{1}{50}\)

    \(=\frac{47}{150}\)

b) 

 \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{19.20}\)

    \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)

    \(=3.\left(1-\frac{1}{20}\right)\)

    \(=3.\frac{19}{20}\)

    \(=\frac{57}{20}\)

TÍNH BẰNG CÁCH NHANH NHẤT NHA CÁC BN 

a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)

b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)

c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)

\(=1-\frac{1}{2021}=\frac{2020}{2021}\)

d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)

e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)

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

\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)

B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101

B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101

B= 1/3 - 1/101

B=98/303

( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )

a, A = 1/2x3+ 1/ 3x4 + 1/4x5 + 1/5x6 + ... + 1/99x100

    A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 + ... + 1/99 -1/100

    A= 1/2 -1/100

    A= 49 / 100

nhờ bạn giải dùm mình câu c,d

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

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

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

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

\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

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

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)

\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)

\(a,M=1-\frac{1}{100}=\frac{99}{100}\)

\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)

                  \(=1-\frac{1}{99}=\frac{98}{99}\)

   =>\(N=\frac{98}{99}:2=\frac{49}{99}\)

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

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

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

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