a: \(P=\left(\dfrac{2x}{\left(2x-3\right)\cdot\left(x-1\right)}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{x-1}\right)\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3x-3-2}{x-1}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-5}=\dfrac{-1}{2x-3}\)
b: Để P>0 thì 2x-3<0
hay x<3/2
c: Để \(P=\dfrac{-1}{x^2-6}\) thì \(x^2-6=2x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
=>(x-3)(x+1)=0
=>x=3 hoặc x=-1