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ta có : x=2010
->x-1=2009
A(x)=x2010-(x-1).x2009 -(x-1).x2008 -...-(x-1).x+1
A(x)=x2010-x2010+x2009-x2009+x2008-...-x2+x+1
A(x)=x+1=2010+1=2011
Ta có:
\(\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)
\(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)
\(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
\(x=-2012\)
ai lam guip toi cau nay voi mai toi nop bai roi
so sanh 2 phan so sau bang cach nahnh nhat: 2007/2008 voi 2008/2009
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\) ( trừ 2 số bằng nhau)
\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
Mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=0-3=-3\)
\(\frac{x+4}{2009}+\frac{x+3}{2010}=\frac{x+2}{2011}+\frac{x+1}{2012}\)\(\Leftrightarrow\)\(\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+1}{2012}+1\right)\)
\(=\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\)
Biểu thức trên chi thỏa mãn khi x+2013=0
\(\Rightarrow x=-2013\)
mk nghĩ là -2013 vì nếu thay x=-2013 vào thì các phân số sẽ bằng -1.
nếu cộng lại thì đc -2
k nhé
\(a,\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
\(\Rightarrow\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)+\left(\frac{x+7}{2008}+1\right)=0\)
\(\Rightarrow\frac{x+2016}{2010}+\frac{x+2016}{2009}+\frac{x+2006}{2008}=0\)
chỉ bt lm v thoi "(
a) \(\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
<=> \(\frac{x+5}{2010}+1+\frac{x+6}{2009}+1+\frac{x+7}{2008}+1=0\)
<=> \(\frac{x+2015}{2010}+\frac{x+2015}{2009}+\frac{x+2015}{2008}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
<=> \(x+2015=0\) (do 1/2010 + 1/2009 + 1/2008 # 0 )
<=> \(x=-2015\)
Vậy...
b) mạo phép chỉnh đề
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+344}{5}=0\)
<=> \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+344}{5}-3=0\)
<=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{5}=0\)
làm tương tự a