Tìm x biết:
\(\frac{x-2}{2015}+\frac{x-3}{2014}=\frac{x-1}{1013}\)
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DT
2
D
28 tháng 3 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2016}-1\right)-\left(\frac{x-3}{2014}-1\right)=\left(\frac{x-4}{2013}-1\right)\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}=\frac{x-2017}{2013}\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(x-2017=0\left(vì\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)
x=2017
21 tháng 4 2015
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
NL
2
IM
10 tháng 9 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 \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
MA
23 tháng 9 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
HC
23 tháng 9 2016
<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)
<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)
<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)
<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)
<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)
<=> \(x=2017\)
Vậy x = 2017
đúng thì
PH
4
21 tháng 6 2017
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
LM
21 tháng 6 2017
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
VT
0
LM
1
TB
12 tháng 4 2017
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:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
=>\(x-2016=0\)
<=>\(x=2016\)
Mk sửa 1013 thành 1008 nhá
\(\frac{x-2}{2015}+\frac{x-3}{2014}=\frac{x-1}{1008}\)
\(\Leftrightarrow\frac{x-2}{2015}+\frac{x-3}{2014}-2=\frac{x-1}{1008}-2\)
\(\Leftrightarrow\left(\frac{x-2}{2015}-1\right)+\left(\frac{x-3}{2014}-1\right)=\frac{x-1}{1013}-2\)
\(\Leftrightarrow\frac{x-2-2015}{2015}+\frac{x-3-2014}{2014}=\frac{x-1-2016}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}=\frac{x-2017}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{1008}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)=0\)
\(\Leftrightarrow x-2017=0\times\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
Hok TOT ^_^