a) Ta có: \(C=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)
\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\left(3-\dfrac{2}{x-1}\right)\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3x-3-2}{x-1}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-5}\)
\(=\dfrac{-\left(3x-5\right)}{\left(3x-5\right)\left(2x-3\right)}=\dfrac{-1}{2x-3}\)
c) Để C>1 thì C-1>0
\(\Leftrightarrow\dfrac{-1}{2x-3}-1>0\)
\(\Leftrightarrow\dfrac{-1-2x+3}{2x-3}>0\)
\(\Leftrightarrow\dfrac{-2x+2}{2x-3}>0\)
\(\Leftrightarrow\dfrac{2x-2}{2x-3}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\2x-2>0\end{matrix}\right.\Leftrightarrow1< x< \dfrac{3}{2}\)