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PN
0
PN
1
G
11 tháng 5 2023
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
TB
1
18 tháng 4 2018
Ta có : \(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+\dfrac{x+3}{2012}+\dfrac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+4}{2011}+1\right)=4\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}=4\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}\right)=4\) \(\Leftrightarrow\left(x+2015\right).0,002=4\) ( mik lấy gần bằng nha )
\(\Leftrightarrow x+2015=2000\Leftrightarrow x=-15\)
Vậy phương trình có nghiệm là x=-15
KE
2
8 tháng 3 2018
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
8 tháng 3 2018
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 ~
NH
1
N
13 tháng 12 2018
\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)
\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)
\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)
\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy x=-2010
24 tháng 2 2022
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
24 tháng 2 2022
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
