Đáp án:
`S=\{-1;3/2\}`
Giải thích các bước giải:
`ĐKXĐ:x\ne -1/2;x\ne±2`
`(2x+1):(2x^2+5x+2)-3:(x^2-4)=2`
`⇔(2x+1)/(2x^2+5x+2)-3/(x^2-4)=2`
`⇔(2x+1)/(2x^2+x+4x+2)-3/(x^2-4)=2`
`⇔(2x+1)/[x(2x+1)+2(2x+1)]-3/(x^2-4)=2`
`⇔(2x+1)/((2x+1)(x+2))-3/(x^2-4)=2`
`⇔1/(x+2)-3/(x^2-4)=2`
`⇔(x-2)/((x-2)(x+2))-3/((x-2)(x+2))=(2(x^2-4))/((x-2)(x+2))`
`⇒(x-2)-3=2(x^2-4)`
`⇔x-5=2x^2-8`
`⇔2x^2-8-x+5=0`
`⇔2x^2-x-3=0`
`⇔2x^2+2x-3x-3=0`
`⇔2x(x+1)-3(x+1)=0`
`⇔(x+1)(2x-3)=0`
\(⇔\left[ \begin{array}{l}x+1=0\\2x-3=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1(TM)\\x=\dfrac{3}{2}(TM)\end{array} \right.\)
Vậy `S=\{-1;3/2\}`
