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pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
Chúc bạn học tốt ~


(x-1)/2015 + x/2014 + 1/503 - (x-3)/2013 - x/2012 - 1/1007 =0
(x-2016)/2015 + (x-2016)/2014 - (x-2016)/2012 - (x-2016)/2013 = 0
(x-2016) ( 1/2015 + 1/2016 - 1/2013 - 1/2012) = 0
Mà 1/2015 + 1/2016 - 1/2013 - 1/2012 khác 0
Suy ra x -2016=0
x=2016
Chỗ nào thắc mắc nhớ hỏi mik nhe!

\(\frac{2-x}{2012}-2=\frac{1-x}{2013}-\frac{x}{1007}\)
\(\Leftrightarrow\frac{2-x}{2012}-\frac{1-x}{2013}+\frac{x}{1007}-2=0\)
\(\Leftrightarrow\left(\frac{2-x}{2012}+1\right)-\left(\frac{1-x}{2013}+1\right)+\left(\frac{x}{1007}-2\right)=0\)
\(\Leftrightarrow\frac{2014-x}{2012}-\frac{2014-x}{2013}+\frac{x-2014}{1007}=0\)
\(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2012}-\frac{1}{2013}-\frac{1}{1007}\right)=0\Leftrightarrow2014-x=0\Leftrightarrow x=2014\)

|x-1|+1=2013.
=>|x-1|=2012.
=>x-1E{-2012;2012}.
=>xE{-2011;2013}(tương ứng).
tk ek nha em mới lớp 6.
-chúc ai tk mk/em học giỏi và may mắn-

Cộng 2 vế với 2 ta có :
5-x^2/2012 + 1 = (4-x^2/2013+1) - (x^2-3/2014-1)
<=> 2017-x^2/2012 = 2017-x^2/2013 - x^2-2017/2014 = 2017-x^2/2013+ 2017-x^2/2014
<=> 2017-x^2/2013 + 2017-x^2/2014 - 2017-x^2/2012 = 0
<=> (2017-x^2).(1/2013+1/2014-1/2012) = 0
<=> 2017-x^2 = 0 ( vì 1/2013+1/2014-1/2012 khác 0 )
<=> x = \(\sqrt{2017}\)
k mk nha
\(\Leftrightarrow\frac{5-x^2}{2012}+1=\frac{4-x^2}{2013}+1+\frac{3-x^2}{2014}+1\)
\(\Leftrightarrow\frac{2017-x^2}{2012}-\frac{2017-x^2}{2013}-\frac{2017-x^2}{2014}=0\)
\(\Leftrightarrow\left(2017-x^2\right)\left(\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow2017-x^2=0\)
\(\Leftrightarrow x^2=2017\)
\(\Leftrightarrow x=\sqrt{2017}\)
V...\(S=\left\{\sqrt{2017}\right\}\)

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+2045}{10}=0\)
\(\Leftrightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1+\frac{x+2045}{10}-3=0\)
\(\Leftrightarrow\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}+\frac{x+2045-3.10}{10}=0\)
\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{10}=0\)
\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\ne0\)
Nên x + 2015 = 0 <=> x = -2015
Vậy x = -2015

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 \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
\(\left|x-1\right|+1=2013\)
\(\Leftrightarrow\left|x-1\right|=2012\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2012\\1-x=2012\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2013\\x=-2011\end{cases}}}\)
Vậy pt có tập nghiệm \(S=\left\{2013;-2011\right\}\)