`x^2-(m-1)x-2=0`
`\Delta=[-(m-1)]^2-4.(-2)`
`\Delta=(m-1)^2+8>0` với `AAm`
Do `\Delta>0` với `AAm` nên pt luôn có 2 nghiệm phân biệt
Theo Viet: `{(x_1+x_2=m-1 (1) ),(x_1.x_2=-2(2)):}`
Có: `x_1/x_2=(x_2^2-3)/(x_1^2-3)`
`<=> x_1(x_1^2-3)=x_2(x_2^2-3)`
`<=> x_1^3-3x_1=x_2^3-3x_2`
`<=> x_1^3-x_2^3-3x_1+3x_2=0`
`<=> (x_1-x_2)(x_1^2+x_1x_2+x_2^2)-3(x_1-x_2)=0`
`<=> (x_1-x_2)[(x_1+x_2)^2-x_1x_2-3]=0`
`-> (x_1-x_2)[(m-1)^2-(-2)-3]=0`
`<=> (x_1-x_2)[(m-1)^2-1]=0`
`<=>`\(\left[ \begin{array}{l}x_1=x_2\\(m-1)^2=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x_1=x_2\\m-1=1\\m-1=-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x_1=x_2 (*)\\m=2\\m=0\end{array} \right.\)
Từ $(1)(*)⇒x_1=x_2=\dfrac{m-1}{2}$
Thay `x_1=x_2=(m-1)/2` vào (2) ta có:
`(m-1)^2/4=-2`
`<=> (m-1)^2=-8` (vô lý)
Vậy `m∈{2;0}`
