1.
\(\dfrac{x+1}{x-2}=\dfrac{1}{x^2-4}\),\(ĐKXĐ,x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+3x+2=1\)
\(\Leftrightarrow x^2+3x+1=0\)
\(\Leftrightarrow\dfrac{-3+\sqrt{9-4.1.1}}{2}=\dfrac{-3+\sqrt{5}}{2}\left(TM\right)\)
Vậy nghiệm PT là: \(x=\dfrac{-3+\sqrt{5}}{2}\)
b.
ta có : \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow-\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\Rightarrow2004-x=0=>x=2014\)
Vậy nghiệm của PT là x = 2014