\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)

Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)

=> x = 2017

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

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

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

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

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

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

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)

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

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

\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)

\(\Rightarrow x+2018=0\Rightarrow x=-2018\)

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

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)

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

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

=> x+2018=0

=> x=-2018

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

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

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

\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)

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

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)

\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)

\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)

\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)

\(\Rightarrow x+4031=0\)

\(\Rightarrow x=-4031\)

\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)

\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)

\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)

=>x+2015=0

<=>x=-2015

=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)

<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)

<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)

<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)

<=> x = -2015

\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(100=2x+4\)

\(\Leftrightarrow\)\(2x=96\)

\(\Leftrightarrow\)\(48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)

\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(49=x+1\)

\(\Leftrightarrow\)\(x=48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

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