Tìm x:
\(\frac{2-x}{2013}-x=\frac{1-x}{2014}-\frac{x}{2015}\)
K
Khách
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Những câu hỏi liên quan
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
HI
2
9 tháng 1 2020
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0
=> x + 2017 = 0
=> x = -2017
9 tháng 1 2020
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)
\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Dễ thấy cái ngoặc to < 0
=> x=-2017
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 \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
MN
1
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:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
=>\(x-2016=0\)
<=>\(x=2016\)
H
2
AM
11 tháng 6 2015
\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)
\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)
=>x+2015=0
<=>x=-2015
11 tháng 6 2015
=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)
<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)
<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)
<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
<=> x = -2015
VN
2
16 tháng 5 2017
Ta có 1+1/2014 +1/x=1/(x+1)+1+1/2013 nên 1/x-1/(x+1)=1/2013-1/(2013+1) nên x=2013
BT
16 tháng 5 2017
Ta có: \(\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2013}-\frac{2015}{2014}\)
<=> \(\frac{1}{x\left(x+1\right)}=\frac{2014^2-2015.2013}{2013.2014}=\frac{1}{2013.2014}\)
<=> x(x+1)=2013.2014
=> x=2013
Đáp số: x=2013
LD
5
11 tháng 8 2015
=> \(\left(\frac{x+4}{2012}+1\right)+\left(\frac{x+3}{2013}+1\right)=\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)
=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
=> x + 2016 = 0 ( Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)
=> x = -2016
\(\Leftrightarrow\dfrac{2-x}{2013}+1-x=\dfrac{1-x}{2014}+1-\dfrac{x}{2015}\)
\(\Leftrightarrow\dfrac{2015-x}{2013}-x=\dfrac{2015-x}{2014}-\dfrac{x}{2015}\)
\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}-\dfrac{x}{2015}+x\)
\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}-\dfrac{x}{2015}+1+x-1\)
\(\Leftrightarrow2015-x=0\)
hay x=2015