Đáp án:

 `S={1;2}.`

Giải thích các bước giải:

 `\frac{3x}{x^{2}+3x+2}+\frac{2x}{x^{2}+x+2}=1` `(đk:x\ne-1;x\ne-2)`

`<=>\frac{3x}{x^{2}+3x+2}-1/2+\frac{2x}{x^{2}+x+2}-1/2=0`

`<=>\frac{6x-(x^{2}+3x+2)}{2.(x^{2}+3x+2)}+\frac{4x-(x^{2}+x+2)}{2.(x^{2}+x+2)}=0`

`<=>\frac{6x-x^{2}-3x-2}{2.(x^{2}+3x+2)}+\frac{4x-x^{2}-x-2}{2.(x^{2}+x+2)}=0`

`<=>\frac{-x^{2}+3x-2}{2.(x^{2}+3x+2)}+\frac{-x^{2}+3x-2}{2.(x^{2}+x+2)}=0`

`<=>1/2 .(-x^{2}+3x-2).(\frac{1}{x^{2}+3x+2}+\frac[1}{x^{2}+x+2})=0`

`<=>-(x^{2}-3x+2).(\frac{x^{2}+x+2+x^{2}+3x+2}{(x^{2}+3x+2).(x^{2}+x+2)})=0`

`<=>-(x^{2}-3x+2).(\frac{2x^{2}+4x+4}{(x^{2}+3x+2).(x^{2}+x+2)})=0`

`+)TH1:-(x^{2}-3x+2)=0`

`<=>x^{2}-x-2x+2=0`

`<=>x.(x-1)-2.(x-1)=0`

`<=>(x-2).(x-1)=0`

`<=>x-2=0` hoặc `x-1=0`

`<=>x=2(tmđk)` hoặc `x=1(tmđk)`

`+)TH2:\frac{2x^{2}+4x+4}{(x^{2}+3x+2).(x^{2}+x+2)}=0`

`=>2x^{2}+4x+4=0`

`<=>2.(x^{2}+2x+2)=0`

`<=>(x^{2}+2x+1)+1=0`

`<=>(x+1)^{2}+1=0`

Do `(x+1)^{2}\ge0AAx=>(x+1)^{2}+1>0AAx`

`=>TH2` Vô nghiệm

Vậy `S={1;2}.`