\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)

\(\Rightarrow x+1=2017\)

\(\Rightarrow x=2017-1=2016\)

Vậy x = 2016

\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2016}{2017}\)

1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{3}{4}\)+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{2013}{8068}\)

Bn tự lm tiếp nhé!!! Sorry mk đang vội

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(x+x+x+x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=1-\frac{15}{16}\)

\(4x=\frac{1}{16}\)

\(x=\frac{1}{16}\div4\)

\(x=\frac{1}{64}\)

ta coi x để ra mội bên ta có 1/2 + 1/4 + 1/8 + 1/6 = 25/24

x + 25/24 = 1

            x = 1 - 25/24

            x = -1/24

mk nha

Nhân lên 16 lần ta có đề bài mới: 

Tìm x biết: \(\left(x+8\right)+\left(x+4\right)+\left(x+2\right)+\left(x+1\right)=16\)

Bài giải

\(x\times4+1+2+4+8=16\)

\(x\times4=1\)

Vậy \(x=\frac{1}{4}\)

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 \(x=2015\)

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

=>1-1/2+1/2-1/3+...+1/x-1/(x+1)=2022/2021

=>1-1/(x+1)=2022/2021

=>1/(x+1)=-1/2021=1/-2021

=>x+1=-2021

=>x=-2022