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 \(x=2015\)

đặt A=.....

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)=\(\frac{2016}{2017}\)

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

=\(\frac{x}{x+1}=\frac{2016}{2017}\)

=>x=2016

vậy..............

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

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

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

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

\(\Rightarrow x+1=2017\)

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

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

(1-1/1.2)+(1-1/2*3)+......+(1-1/2015*2016)

=(0/1*2)+(0+2*3)+..........+(0/2015*2016)

=0

tui nghĩ cái đề phải như thế này  \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)

=0 con minh ko giai

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)

\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)

\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)

(x + 1/1.2) + (x + 1/2.3) + (x + 1/3.4) + ... + (x + 1/2022.2023) = 2023x

x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x

2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x

2023x - 2022x = 1 - 1/2023

x = 2022/2023