\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)

\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)

\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)

\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

Bạn ơi ghi sai đề rồi kìa

CÓ: \(\frac{x-1}{2015}+\frac{x-2}{2014}-\frac{x-3}{2013}-\frac{x-4}{2012}=0\)\(0\)

<=>\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)=0\)

<=>\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)

<=>\(\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Do:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)

=>\(x-2016=0\)

<=>\(x=2016\)

x=- 2016

\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)

\(\Leftrightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)

\(\frac{x+5+2012}{2012}+\frac{x+4+2013}{2013}=\frac{x+3+2014}{2014}+\frac{x+2+2015}{2015}\)

\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)

\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)

\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)

\(\Rightarrow x+2017=0\)

\(\Rightarrow x=-2017\)

\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)

\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)

\(\left(x+2017\right)\cdot\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\)

Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)

suy ra \(x+2017=0\)

suy ra  \(x=-2017\)

Vậy   \(x=-2017\)

Ta có:

\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)

\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)

\(\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+4}{2013}+\frac{2013}{2013}=\frac{x+3}{2014}+\frac{2014}{2014}+\frac{x+2}{2015}+\frac{2015}{2015}\)

\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)

\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)

\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

\(=>\orbr{\begin{cases}x+2017=0\\\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}=0\end{cases}}\)

Mà:   \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)

=> x + 2017 = 0 => x = -2017

có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1

=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)

vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015

x=2015

=> \(\left(\frac{x+4}{2012}+1\right)+\left(\frac{x+3}{2013}+1\right)=\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)

=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

=> x + 2016 = 0 ( Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)

=> x = -2016

cộng mỗi phân số cho 1  

\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)

\(\Leftrightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

\(\Leftrightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)

\(\Leftrightarrow\left(x+2006\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

\(\Rightarrow x+6=0\)  ( vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\)  khac 0 )

=> x = -6 

Bạn xem lại đề nhé U Suck

\(\frac{x+1}{2015}\)+\(\frac{x+2}{2014}\)+\(\frac{x+3}{2013}\)+\(\frac{x+4}{2012}\)=44

\(\frac{x+1}{2015}\)+1+\(\frac{x+2}{2014}\)+1+\(\frac{x+3}{2013}\)+1+\(\frac{x+4}{2012}\)+1=44+4

\(\frac{x+2016}{2015}\)+\(\frac{x+2016}{2014}\)+\(\frac{x+2016}{2013}\)+\(\frac{x+2016}{2012}\)=48

(x+2016)(\(\frac{1}{2015}\)+\(\frac{1}{2014}\)+\(\frac{1}{2013}\)+\(\frac{1}{2012}\))=48

tu lam tiep.Nho k tui voi

KẾT QUẢ =4 

XIN LỖI NHA

\(\Rightarrow\frac{x+1}{2015}=1;\frac{x+2}{2014}=1;\frac{x+3}{2013}=1;\frac{x+4}{2012}=1\)

\(vì\)\(1+1+1+1=4\)

\(\text{mUỐN PHÂN SÓ =1 THÌ TỬ SỐ = MẪU SỐ}\)

\(X+1=2015\)

\(X=2014\)

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