a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(A=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-2}{2\left(x-1\right)}+\dfrac{3}{2\left(x-1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right):\dfrac{x+1-x+3}{x+1}\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{x+1}{4}\)
\(=\dfrac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{4}\)
\(=\dfrac{x^2+2x+1-\left(x^2+2x-3\right)}{2\left(x-1\right)}\cdot\dfrac{1}{4}\)
\(=\dfrac{4}{2\left(x-1\right)\cdot4}=\dfrac{1}{2\left(x-1\right)}\)
Thay x=2005 vào A, ta được:
\(A=\dfrac{1}{2\left(2005-1\right)}=\dfrac{1}{4008}\)
b: A=-1002
=>\(\dfrac{1}{2\left(x-1\right)}=-1002\)
=>2(x-1)=-1/1002
=>\(x-1=-\dfrac{1}{2004}\)
=>\(x=\dfrac{2003}{2004}\left(nhận\right)\)
