ĐKXĐ : \(x\ne\dfrac{1}{2}\)
\(P=\dfrac{2x^2+3x+3}{2x-1}=\dfrac{x\left(2x-1\right)+2\left(2x-1\right)+5}{2x-1}=\dfrac{\left(x+2\right)\left(2x-1\right)+5}{2x-1}=x+2+\dfrac{5}{2x-1}\)
Vì \(x\in Z\Rightarrow x+2\in Z\)
=> Để \(P\in Z\) thì \(\dfrac{5}{2x-1}\in Z\)
\(\Leftrightarrow5⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1;5;-5\right\}\)
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