Đáp án:
`S=\{0\}`
Giải thích các bước giải:
`ĐKXĐ:x\ne -3;x\ne 1`
`\frac{x}{2(x+3)}+\frac{x}{2x-2}=\frac{2x}{(x-1)(x+3)}`
`<=>\frac{x}{2(x+3)}+\frac{x}{2(x-1)}=\frac{2x}{(x-1)(x+3)}`
`<=>\frac{x(x-1)}{2(x+3)(x-1)}+\frac{x(x+3)}{2(x-1)(x+3)}=\frac{2x.2}{2(x-1)(x+3)}`
`=>x(x-1)+x(x+3)=2x.2`
`<=>x^2-x+x^2+3x=4x`
`<=>2x^2+2x=4x`
`<=>2x^2+2x-4x=0`
`<=>2x^2-2x=0`
`<=>2x(x-1)=0`
`TH1:2x=0<=>x=0(TM)`
`TH2:x-1=0<=>x=1(KTM)`
Vậy `S=\{0\}`
