x-2/2016 + x-3/2017 + x-4/2018 +3=0
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CE
4
23 tháng 3 2019
\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\\ \Leftrightarrow\left(\frac{x-2}{2016}+1\right)+\left(\frac{x-3}{2017}+1\right)+\left(\frac{x-4}{2018}+1\right)=0\\ \Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\\ \Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\\ Vì\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\ne0\\ \Rightarrow x+2014=0\\ \Leftrightarrow x=-2014\\ Vậy...\)
NH
1
19 tháng 2 2020
Ta có\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\)
\(\Leftrightarrow\frac{x-2}{2016}+1+\frac{x-3}{2017}+1+\frac{x-4}{2018}=0\)
\(\Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\) Vì \(\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)>0\)
\(\Rightarrow x+2014=0\)
\(\Rightarrow x=-2014\)
PN
3
DS
0
PH
6
5 tháng 3 2017
Bài này có quy tắc đấy
Mình sẽ gúp cậu nhưng sai thì thôi nhé
DT
0
TN
5 tháng 7 2016
x+2015/5 + x+2016/4=x+2017/3 + x+2018/2
\(\Rightarrow\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Rightarrow x+2020=0\).Do \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x=-2020\)
\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\)
\(\Leftrightarrow\left(\frac{x-2}{2016}+1\right)+\left(\frac{x-3}{2017}+1\right)+\left(\frac{x-4}{2018}+1\right)=0\)
\(\Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\ne0\)
\(\Leftrightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
Vậy \(x=-2014\)