ĐK: $\left[\begin{array}{l}x>1\\\dfrac{1}{2}<x<1\\x\ne\dfrac{3}{2}\end{array}\right.$
Bất phương trình tương đương với:
$\dfrac{-2}{\log_3(2x^2-3x+1)}\ge\dfrac{-1}{\log_3(x+1)}$
$\Leftrightarrow \dfrac{2}{\log_3(2x^2-3x+1)}\le\dfrac{1}{\log_3(x+1)}$
$\Leftrightarrow \left[\begin{array}{l}\log_3(2x^2-3x+1)\ge2\log_3(x+1)\\\left\{\begin{array}{l}\log_3(2x^2-3x+1)<0\\\log_3(x+1)>0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x^2-3x+1\ge(x+1)^2\\\left\{\begin{array}{l}0<2x^2-3x+1<1\\x+1>1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge5\\x<0\\0<x<\dfrac{1}{2}\\1<x<\dfrac{3}{2}\end{array}\right.$
Kết hợp điều kiện ta được: $x\in(0;\dfrac{1}{2})\cup(1;\dfrac{3}{2})\cup[5;+\infty)$