`(x+1)/(x-2)-(x-1)/(x+2)=(2(x^2+2))/(x^2-4)` ĐKXĐ: `x\ne2;x\ne-2`
`<=>frac{(x+1)(x+2)}{(x-2)(x+2)}-frac{(x-1)(x-2)}{(x+2)(x-2)}=frac{2x^2+4}{(x-2)(x+2)}`
`=>(x+1)(x+2)-(x-1)(x-2)=2x^2+4`
`<=>x^2+2x+x+2-(x^2-2x-x+2)=2x^2+4`
`<=>x^2+3x+2-x^2+3x-2=2x^2+4`
`<=>6x=2x^2+4`
`<=>2x^2-6x+4=0`
`<=>2x^2-2x-4x+4=0`
`<=>(2x^2-2x)-(4x-4)=0`
`<=>2x(x-1)-4(x-1)=0`
`<=>(x-1)(2x-4)=0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\2x-4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\2x=4\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1(TMĐK)\\x=2(KTMĐK)\end{array} \right.\)
Vậy phương trình trên có nghiệm `S={1}`
