`a)` Với `x \ne +-3;x \ne -1` có:
`Q=([2x]/[x+3]+x/[x-3]-[3x^2+3]/[x^2-9]):([2x-2]/[x-3]-1)`
`Q=[2x(x-3)+x(x+3)-3x^2-3]/[(x-3)(x+3)]:[2x-2-x+3]/[x-3]`
`Q=[2x^2-6x+x^2+3x-3x^2-3]/[(x-3)(x+3)].[x-3]/[x+1]`
`Q=[-3x-3]/[(x+3)(x+1)]=[-3(x+1)]/[(x+3)(x+1)]=[-3]/[x+3]`
`b)` Với `x \ne +-3;x \ne -1` có:
`Q < [-1]/3<=>[-3]/[x+3] < [-1]/3`
`<=>[-3]/[x+3]+1/3 < 0`
`<=>[-9+x+3]/[3(x+3)] < 0`
`<=>[x-6]/[x+3] < 0`
`@TH1:{(x-6 < 0),(x+3 > 0):}<=>{(x < 6),(x > -3):}<=>-3 < x < 6`
Mà `x \ne +-3;x \ne -1`
`=>-3 < x < 6;x \ne 3;x \ne -1`
`@TH2:{(x-6 > 0),(x+3 < 0):}<=>{(x > 6),(x < -3):}=>` (Vô lí)
Vậy `-3 < x < 6;x \ne -3;x \ne -1`
