\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}=\frac{2013}{2014}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2013}{2014}\)

\(\Rightarrow1-\frac{1}{n+1}=\frac{2013}{2014}\)

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

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

\(\Rightarrow n+1=2014\)

\(\Rightarrow n=2014-1\)

\(\Rightarrow n=2013\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

\(=\dfrac{24}{25}\)

Nhanh giúp mình với cả nhà ơi

a)(y+2):5-5x5=378

(y+2):5-25=378

(y+2):5=378+25

(y+2):5=403

(y+2)=403x5

y+2=2015

y=2015-2

y=2013

(y+2):5-5.5=378

(y+2):5-25=378

(y+20)=378+25

(y+2)=403

(y+2)=403.5

y+2=2015

y=2015-2

y=2013

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

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

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

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

=> x+1 = 100

=> x = 100 - 1 

=> x = 99

mơ đi Nguyễn Đình Dũng

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

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

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

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

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

\(\Rightarrow x=16\)

?

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

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

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

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

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

=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100

=1-1/100

=99/100

e ko cop đâu nhé e lớp 6 câu nay e làm đc ạ !

`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

Ta có : A = \(\frac{1}{1\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{3\text{x}4}+...+\frac{1}{X\text{x}\left(X+1\right)}\)

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

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

           A = \(\frac{x}{x+1}\)

Ủng hộ mik nhá !!!!

Ta có:

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

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

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

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

\(\Rightarrow x+1=?\Leftrightarrow x=?\)