\(\dfrac{x}{x+1}-\dfrac{x-1}{x-3}=\dfrac{4}{x^2-2x-3}\)
ĐK: \(x\ne-1;x\ne3\)
\(\Leftrightarrow\dfrac{x\left(x-3\right)}{x^2-2x-3}-\dfrac{x^2-1}{x^2-2x-3}=\dfrac{4}{x^2-2x-3}\)
\(\Leftrightarrow x^2-3x-x^2+1=4\)
\(\Leftrightarrow-3x=3\)
\(\Leftrightarrow x=-1\left(KTMĐK\right)\)
PTVN