a) Thay k=2 vào phương trình, ta được:
\(\dfrac{x+2}{x+3}+\dfrac{x-3}{x-1}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-1\right)+\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{2\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(x^2-x+2x-2+x^2-9-2x^2-4x+6=0\)
\(\Leftrightarrow-3x-5=0\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)

