
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


(x-1)/2016 +(x-2)/2015 -(x-3)/2014 = (x-4)/2013. =>(x-1)/2016 +(x-2)/2015 = (x-3)/2014 + (x-4)/2013. =>. (X-1)/2016 -1 + (x-2)/2015 -1 = (x -3)/2014 -1 + (x-4)/2013 -1 => (x -2017)/2016 + (x-2017)/2015 -(x-2017)/2014 -(x-2017)/2013 =0. => (x-2017)(1/2016 +1/2015 -1/2014 -1/2013) = 0 => x-2017 =0 => x = 2017
Ta có: \(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\Leftrightarrow\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}-\frac{x-4}{2013}=0\)
\(\Leftrightarrow\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)

c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)

\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\left(\frac{x-1}{2016}+1\right)+\left(\frac{x-2}{2015}+1\right)=\left(\frac{x-3}{2014}+1\right)+\left(\frac{x-4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2014}\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x+2017}{2014}-\frac{x+2017}{2013}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)

\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016} \)
\(\Leftrightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1-\frac{x+2}{2015}-1-\frac{x+1}{2016}-1=0\)
\(\Leftrightarrow\frac{x+2014+2013}{2013}+\frac{x+3+2014}{2014}-\frac{x+2+2015}{2015}-\frac{x+1+2016}{2016}=0\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2017=0\) ( vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\)>0)
\(\Leftrightarrow x=2017\)

C1:
b)=>y-1,5=0
=>y=1,5
(x-1)2=0
=>1
Vậy x=1;y=1,5
Ai thấy đúng thì


\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}\)
\(\Rightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1\)
\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}\)
\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
\(Do\)\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
Vậy \(x=-2017\)
bạn bấm vào "đúng 0" là sẽ có đáp án hiện ra
vì 2>0\(\Rightarrow|x-2014|+|x-2015|+|x-2016|>0\)
\(\Rightarrow|x-2014|+|x-2015|+|x-2016|\)
\(\Rightarrow x-2014+x-2015+x-2016=2\)
\(\Rightarrow x+x+x-2014-2015-2016=2\Rightarrow3x-6045=2\)
\(\Rightarrow3x=6047\Rightarrow x=6047:3=\frac{6047}{3}\)
bạn jj vừa trả lời ơi, cho mik hỏi tí là vì sao bn suy ra đc dòng 3