Đáp án+Giải thích các bước giải:
`c) 7/(2x^2-x-1)+1/(2x+1)=6/(x^2-1)` `ĐKXĐ: x` $\neq$ `±1;` `x`$\neq$ `(-1)/2`
`⇔7/(2x^2-2x+x-1)+1/(2x+1)=6/((x-1)(x+1))`
`⇔7/(2x(x-1)+(x-1))+1/(2x+1)=6/((x-1)(x+1))`
`⇔7/((2x+1)(x-1))+1/(2x+1)=6/((x-1)(x+1))`
`⇔(7(x+1))/((2x+1)(x-1)(x+1))+((x-1)(x+1))/((2x+1)(x-1)(x+1))=(6(2x+1))/(2x+1)(x-1)(x+1)`
⇒`7(x+1)+(x-1)(x+1)=6(2x+1)`
⇔`7x+7+x^2-1-12x-6=0`
⇔`x^2-4x=0`
⇔`x(x-4)=0`
⇔$\left[\begin{matrix} x=0\\ xx-4=0\end{matrix}\right.$⇔$\left[\begin{matrix} x=0(TM)\\ x=4(TM)\end{matrix}\right.$
Vậy `S={0;4}`