`a)`

Thay `m=1`, ta có:

`x^2-x-1-1=0`

`x^2-x-2=0`

`x^2-2x+x-2=0`

`x(x-2)+(x-2)=0`

`(x+1)(x-2)=0`

`x+1=0` hoặc `x-2=0`

`x=-1` hoặc `x=2`

Vậy `S={-1;2}`

`2)`

`x^2-x-m-1=0`

`Δ=(-1)^2-4.1.(-m-1)=1+4m+4=4m+5`

Để pt có `2` nghiệm pb thì `4m+5>0=>m> -5/4`

Theo viète, ta có:

`x_1+x_2=-b/a=1`

`x_1x_2=c/a=-m-1`

Ta có: `x_1^2+x_1x_2-6x_2^2=0`

`(x_1^2+x_1x_2-6x_2^2)/(x_2^2)=0`

`((x_1)/(x_2))^2+(x_1)/(x_2)-6=0`

Đặt `a=(x_1)/(x_2)`, ta có:

`a^2+a-6=0`

`a^2+3a-2a-6=0`

`a(a+3)-2(a+3)=0`

`(a-2)(a+3)=0`

`a-2=0` hoặc `a+3=0`

`a=2` hoặc `a=-3`

Th1: `a=2=>(x_1)/(x_2)=2=>x_1=2x_2`

Ta có: `x_1+x_2=1`

`2x_2+x_2=1`

`3x_2=1`

`x_2=1/3`

`=>x_1=2. 1/3=2/3`

Ta có: `x_1x_2=-m-1`

`1/3 . 2/3=-m-1`

`-m-1=2/9`

`-m=11/9`

`m=-11/9 (tm)`

Th2: `a=-3=>(x_1)/(x_2)=-3=>x_1=-3x_2`

Ta có: `x_1+x_2=1`

`-3x_2+x_2=1`

`-2x_2=1`

`x_2=-1/2`

`=>x_1=-3 . (-1/2)=3/2`

Ta có: `x_1x_2=-m-1`

`-1/2 . 3/2=-m-1`

`-m-1=-3/4`

`-m=1/4`

`m=-1/4 (tm)`

Vậy `m in {-11/9;-1/4}`