Question

Cho Q= (2x/x+3+x/x-3-3x^2+3/x^2-9) : (2x-2/x-3-1)
a, Rút gọn Q
b, Tìm x để Q< -1/3

Answer 1

`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`