c)$\dfrac{2}{x-1}$ +$\dfrac{2x+3}{x^2+x+1}$ =$\dfrac{(2x-1)(2x+1)}{x^3-1}$
` MC:x^3-1`
`ĐKXĐ:xne1`
⇔$\dfrac{2(x^2+x+1)}{(x-1)(x^2+x+1)}$ +$\dfrac{(2x+3)(x-1)}{(x-1)(x^2+x+1)}$ =$\dfrac{(2x-1)(2x+1)}{x^3-1}$
⇒`2x^2+2x+2+2x^2-2x+3x-3=4x^2-1`
⇔`2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0`
⇔`3x=0`
⇔`x=0(nhận)`
`Vậy S={0}`
f)$\dfrac{1}{x-1}$ +$\dfrac{2x^2-5}{x^3-1}$ =$\dfrac{4}{x^2+x+1}$
` MC:x^3-1`
`ĐKXĐ:xne1`
⇔$\dfrac{x^2+x+1}{(x-1)(x^2+x+1)}$ +$\dfrac{2x^2-5}{x^3-1}$ =$\dfrac{4(x-1)}{(x-1)(x^2+x+1)}$
⇒`x^2+x+1+2x^2-5=4x-4`
⇔`x^2+x+1+2x^2-5-4x+4=0`
⇔`3x^2-3x=0`
⇔`3x(x-1)=0`
⇔$\left \{ {{3x=0} \atop {x-1=0}} \right.$
⇔$\left \{ {{x=0(nhận)} \atop {x=1(loại)}} \right.$
`Vậy S={0}`
