`(x-1)/5 + (x-2)/6 = (x-3)/7 + (x-4)/8`
`⇔ ( (x-1)/5 + 1 ) + ( (x-2)/6 + 1 ) = ( (x-3)/7 + 1) + ( (x-4)/8 + 1)`
`⇔ ( (x-1)/5 + 5/5 ) + ( (x-2)/6 + 6/6 ) = ( (x-3)/7 + 7/7 ) + ( (x-4)/8 + 8/8 )`
`⇔ (x+4)/5 + (x+4)/6 = (x+4)/7 + (x+4)/8`
`⇔ (x+4)/5 + (x+4)/6 - (x+4)/7 - (x+4)/8 = 0`
`⇔ (x+4)(1/5 + 1/6 - 1/7 - 1/8) = 0`
Vì `1/5 + 1/6 - 1/7 - 1/8 \ne 0`
`⇔ x + 4 = 0`
`⇔ x = -4`
Vậy `S = {-4}`
