Lời giải:

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x(x+1)}=\frac{2014}{2015}$

$\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}=\frac{2014}{2015}$

$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}$

$1-\frac{1}{x+1}=\frac{2014}{2015}$

$\frac{1}{x+1}=1-\frac{2014}{2015}=\frac{1}{2015}$
$\Rightarrow x+1=2015$

$\Rightarrow x=2014$

\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{2}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(x+1=20\)

\(x=20-1\)

\(x=19\)

Có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(\Rightarrow x+1=20\Leftrightarrow x=19\)

.3-2/2.3 + 4-3/3.4 + 5-4/4.5 + 6-5/5.6 +...+ 20-19/19.20=18/x

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +...+ 1/19 - 1/20=18/x

1/2 - 1/20=18/x

10/20 - 1/20=18/x

9/20=18/x

18/40=18/x

=>x=40

Vậy x=40

Thanks

1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2

1+2.(1/2-1/x+1)=3/2

1-2/x+1=3/2-1

tự tính

Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2001}\)

\(\Rightarrow\)x+1=2001

x=2000

Vậy x=2000.