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

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

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

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

 Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)

\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)

\(\Rightarrow x+1=2014\)

\(\Rightarrow x=2014-1\)

\(\Rightarrow x=2013\)

Vậy x=2013

 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)

\(1-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\frac{1}{x+1}=1-\frac{2013}{2014}\)

\(\frac{1}{x+1}=\frac{1}{2014}\)

Vì \(x+1\)là mẫu số nên:

\(x+1=2014\)

\(x=2014-1=2013\)

Vậy ....

  P/s: Dấu . là nhân nha!

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

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

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

=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

=> x+1 = 100

=> x = 100 - 1 

=> x = 99

mơ đi Nguyễn Đình Dũng

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{299}{600}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{600}\)

\(\Rightarrow x+1=600\)

\(\Rightarrow x=600-1\)

\(\Rightarrow x=599\)

\(Vậy\) \(x=599\)

`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`

`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`

`=>x/(x+1)=1-1/32`

`=>x/(x+1)=31/32`

`=>32x=31(x+1)`

`=>32x=31x+31`

`=>32x-31x=31`

`=>x=31`

31/31

làm ơn hãy giúp mình

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{103}{50}\right)\right]\cdot2=89\)

\(\left(1-\frac{1}{10}\right)\cdot100-\frac{5}{2}:\left(x+\frac{103}{50}\right)\cdot2=89\)

\(\frac{9}{10}\cdot100-\frac{5}{2}\cdot2:\left(x+\frac{103}{50}\right)=89\)

\(90-5\cdot\left(x+\frac{103}{50}\right)=89\)

\(5\cdot\left(x+\frac{103}{50}\right)=1\)

\(x+\frac{103}{50}=\frac{1}{5}\)

\(x=-\frac{93}{50}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{\left(x-1\right)\times x}=\dfrac{15}{16}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{15}{16}\)

\(1-\dfrac{1}{x}=\dfrac{15}{16}\)

\(\dfrac{1}{x}=1-\dfrac{15}{16}=\dfrac{16}{16}-\dfrac{15}{16}\)

\(\dfrac{1}{x}=\dfrac{1}{16}\)

\(\Rightarrow x=16\)

?

=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2

=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)

Với x+2020=0=>x=-2020

Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí 

Vậy x=-2020