\(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)

\(\Leftrightarrow\frac{x+1}{2017}+1+\frac{x+2}{2016}+1=\frac{x+3}{2015}+1+\frac{x+4}{2014}+1\)

\(\Leftrightarrow\frac{x+2018}{2017}+\frac{x+2018}{2016}-\frac{x+2018}{2015}-\frac{x+2018}{2014}=0\)

\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\right)=0\Leftrightarrow x=-2018\)

Giúp gì mới được 

ủa giúp gì vậy bạn

giup gi

S₁ = 1-2+3-4+....+2017-2018

     = (-1)+(-1)+....+(-1)

     = (-1) x 1009

     =   -1009

S3=2019 nha, mình ko kip viết cách giai

S₁= (1-2)+(3-4)+............+(2017-2018)

     =(-1).(-1).......(-1)

     =(-1).1009

     (-1009)

\(B=\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)

\(B=2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)

\(B=1+\left(\frac{2015}{2}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)\)

\(B=\frac{2017}{2017}+\frac{2017}{2}+...+\frac{2017}{2015}+\frac{2017}{2016}\)

\(B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\frac{B}{A}=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{2}{2017}}=2017\)

kết quả phép tính này =0 nha bạn

k mik nha.

2A = 1+1/2+1/2^2+.....+1/2^2015

A = 2A - A = (1+1/2+1/2^2+......+1/2^2015)-(1/2+1/2^2+.....+1/2^2016)

                 = 1 - 1/2^2016

Tk mk nha

\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\)
\(\Rightarrow A-\frac{1}{2}A=\frac{1}{2}-\frac{1}{2^{2017}}\)
\(\Leftrightarrow A=1-\frac{1}{2^{2016}}\)