`#Ly`

`(4x)/(4x-3)+5/(3x-4x^2)=(x+2)/x`                `(đk : x \ne 3/4 ; x\ne 0)`

`<=>(x . 4x)/(x(4x-3)) - 5/(4x^2-3x)=((x+2)(4x-3))/(x(4x-3))`

`<=>(4x^2)/(4x^2-3x)-5/(4x^2-3x)=(4x^2+5x-6)/(4x^2-3x)`

`<=>(4x^2)/(4x^2-3x)-5/(4x^2-3x)-(4x^2+5x-6)/(4x^2-3x)=0`

`<=>(4x^2-5-4x^2-5x+6)/(4x^2-3x)=0`

`<=>(-5x+1)/(4x^2-3x)=0`

`<=>-5x+1=0`

`<=>-5x=-1`

`<=>x=1/5` `(tm)`

Vậy `S={1/5}`