`[2x]/[x+3]-[x-1]/[3-x]=[3x^2+1]/[x^2-9]` `ĐK: x \ne +-3`
`<=>[2x(x-3)+(x-1)(x+3)]/[(x-3)(x+3)]=[3x^2+1]/[(x-3)(x+3)]`
`=>2x^2-6x+x^2+3x-x-3=3x^2+1`
`<=>-4x=4`
`<=>x=-1` (t/m)
Vậy `S={-1}`
Đúng 4
Bình luận (0)