ĐKXĐ : ` x\ne 0 ; x \ne 6`
` 90/x -36/(x-6) =2`
` => (90(x-6))/(x(x-6)) - (36x)/(x(x-6)) = ( 2x(x-6))/(x(x-6)`
` => (90x-540-36x-2x^2+12x)/(x(x-6)) =0`
` => 90x-540-36x-2x^2+12x =0`
` => -2x^2 + 66x -540 = 0`
` => -x^2 + 33x -270 = 0`
` => x^2 -33x +270 = 0`
` => (x-18)(x-15) =0`
` =>` \(\left[ \begin{array}{l}x=18\\x=15\end{array} \right.\)
Vậy ` x \in { 15;18}`