`3/(1 - 4x)` = `2/(4x + 1)` - `(8 + 6x)/(16x^2 - 4)` ( đk x $\neq$ ± `1/4`)
⇔ `3/(4x - 1)` = `2/(4x + 1)` - `(8 + 6x)/((4x+1)(4x-1))`
⇔`2/(4x+1)` + `3/(4x - 1)` - `(8 + 6x)/((4x+1)(4x-1))` = 0
⇔`(2(4x -1))/((4x+1)(4x-1))` + `(3(4x + 1))/((4x+1)(4x-1))` - `(8 + 6x)/((4x+1)(4x-1))` = 0
⇔`(8x - 2 + 12x + 3 - 8 - 6x )/((4x+1)(4x-1))` = 0
⇔`(14x -7)/((4x+1)(4x-1))`=0
`Do (4x+1)(4x-1) \ne 0 `
`⇒14x - 7 = 0`
`⇒ x = 1/2` ( tm)
Vậy x = `1/2`
