`(2x-3)/12 = 3/(2x-3)`
`<=> (2x-3)(2x-3) = 3.12`
`<=> (2x-3)^2 = 36`
`<=> (2x-3)^2 = (±6)^2`
`<=>` \(\left[ \begin{array}{l}2x-3=6\\2x-3=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=9\\2x=-3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4,5\\x=-1,5\end{array} \right.\)
Vậy `x∈{4,5;-1,5}`