`\color{#FF0000}{T}\color{#FF7F24}{i}\color{#FFFF00}{t}\color{#7FFF00}{u}\color{#00F5FF}{u}\color{#FF1493}{u}`

`x^2-(m-1)x-3m-6=0`

Có `\Delta=(m-1)^2-4(-3m-6)`

`=m^2-2m+1+12m+24`

`=m^2+10m+25`

`=(m+5)^2>=0 AA m`

`->` Pt có `2` nghiệm `AA m`

Theo Viét có: `x_1+x_2=m-1; x_1x_2=-3m-6`

Có `|x_1|=5+|x_2|`

`-> (|x_1|-|x_2|)^2=25`

`-> (x_1+x_2)^2-2x_1x_2-2|x_1x_2|-25=0`

`-> (m-1)^2-2(-3m-6)-2|-3m-6|-25=0`

`-> m^2-2m+1+6m+12-2|3m+6|-25=0`

`-> m^2+4m-12-2|3m+6|=0`

`_` `TH1: m>= -2 -> |3m+6|=3m+6`

`-> m^2+4m-12-6m-12=0`

`-> m^2-2m-24=0`

`-> (m-6)(m+4)=0`

`-> m=6 (TM)` hoặc `m=-4 (KTM)`

`_` `TH2: m< -2 -> |3m+6|=-3m-6`

`-> m^2+4m-12+6m+12=0`

`-> m^2+10m=0`

`-> m(m+10)=0`

`-> m=0 (KTM)` hoặc `m=-10 (TM)`

Vậy `m in {-10; 6}`