$P=1-3x + \dfrac{3}{2-x}\\ =6-3x + \dfrac{3}{2-x}-5\\ =3(2-x)+ \dfrac{3}{2-x}-5\\ t=2-x\\ x<2 \Rightarrow t>0\\ P=3t+ \dfrac{3}{t}-5\\ \ge 2\sqrt{3t.\dfrac{3}{t}}-5(Cauchy)\\ =1$
Dấu "=" xảy ta $\Leftrightarrow \left\{\begin{array}{l} 3t=\dfrac{3}{t} \\t>0\end{array} \right. \Leftrightarrow t=1 \Rightarrow x=1$
Vậy $min_P=1$ khi $x=1$