Question

[vung nguyen thi]

Giải phương trình:

a/ \(\dfrac{x+1}{x^2+x+1}\) - \(\dfrac{x-1}{x^2-x+1}\) = \(\dfrac{3}{x\left(x^4+x^2+1\right)}\)

b/ \(\dfrac{9-x}{2009}\) + \(\dfrac{11-x}{2011}\) = 2

c/ \(\dfrac{15-x}{2010}\) + \(\dfrac{17-x}{2012}\) + \(\dfrac{19-x}{2014}\) = 3

d/ \(\dfrac{x-2014}{2007}\) + \(\dfrac{x-2012}{2009}\) + \(\dfrac{x-10}{2011}\) = \(\dfrac{x-2017}{2014}\) + \(\dfrac{x-2009}{2012}\) + \(\dfrac{x-2011}{2010}\)

Answer 1

[Chí Cường]

a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)

ĐK:\(x\ne0\)

\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)

Answer 2

[Chí Cường]

\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)