`1/(x-1) + 1/((x-1)(x-2)) = 3/(2x)`
ĐKXĐ : `x \ne 1 , x \ne 2`
`⇔ (2x(x-2))/(2x(x-1)(x-2)) + (2x)/(2x(x-1)(x-2)) = (3(x-1)(x-2))/(2x(x-1)(x-2))`
`⇔ 2x(x-2) + 2x = 3(x-1)(x-2)`
`⇔ 2x^2 - 2x = 3x^2 - 9x + 6`
`⇔ 3x^2 - 7x + 6 = 2x^2`
`⇔ x^2 - 7x + 6 = 0`
`⇔ (x-6)(x-1) = 0`
`⇔`\(\left[ \begin{array}{l}x-6=0\\x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=6(TM)\\x=1(KTM)\end{array} \right.\)
Vậy `S = {6}`
