Lời giải:

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

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

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

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

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

$\Rightarrow x=2014$

1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016

<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016

<=> 1 - 1/x+1 = 2015/2016

<=> 1/x+1 = 1/2016

<=> x + 1 = 2016

<=> x = 2015

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

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

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)

 \(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2016}\)

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

\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(x=2016-1=2015\)

            Đáp số: 2015

Hình như bn viết sai đề,là 1/x.(x+1) chứ

ukm mik xin lỗi mik viết sai đề đó

Đề sai nhé phải là x(x+1)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\Leftrightarrow\frac{x}{x+1}=\frac{2015}{2016}\Rightarrow x=2015\)

Vậy \(x=2015\)

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(x-1\right)x}=2\)

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

suy ra \(1-\frac{1}{x}=2\)

hay \(\frac{x-1}{x}=2\) .suy ra x-1=2x .tính ra ta có x=-1