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

=\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2106}{2013}-\frac{x-2016}{2012}\)

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

Mà: \(\frac{1}{2012}>\frac{1}{2015}\)  và \(\frac{1}{2014}< \frac{1}{2013}\)

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

Nên: \(x-2016=0\)

=>\(x=2016\)

Theo bài ra , ta có :

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

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

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

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

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

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

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

\(\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

Vậy \(x=-2016\)

Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)

Chúc bạn học tốt =)) 

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

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

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

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

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

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

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

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

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

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

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

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

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

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

=>\(x-2016=0\)

<=>\(x=2016\)

cho mk hỏi cách giải bài đó đi đáp án mk pk rồi

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

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

\(vì\)\(1+1+1+1=4\)

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

\(X+1=2015\)

\(X=2014\)

tƯƠNG TỰ CZZZZZZZZZZZZ

x=- 2016

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

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

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

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

Vậy x = 2012

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10