\(x^2-\left(m-1\right)x-2=0\)

a=1; b=-m+1; c=-2

Vì a*c=-2<0

nên phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left[-\left(m-1\right)\right]}{1}=m-1\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-2}{1}=-2\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=\left(m-1\right)^2-4\cdot\left(-2\right)=\left(m-1\right)^2+8\)

=>\(x_1-x_2=\pm\sqrt{\left(m-1\right)^2+8}\)

\(\dfrac{x_1}{x_2}=\dfrac{x_2^2-3}{x_1^2-3}\)

=>\(x_1\left(x_1^2-3\right)=x_2\left(x_2^2-3\right)\)

=>\(x_1^3-x_2^3=3x_1-3x_2\)

=>\(\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1x_2-3\right)=0\)

=>\(\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2-3\right]=0\)

=>\(\left[{}\begin{matrix}x_1-x_2=0\\\left(m-1\right)^2-\left(-2\right)-3=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\sqrt{\left(m-1\right)^2+8}=0\left(vôlý\right)\\\left(m-1\right)^2-1=0\end{matrix}\right.\)

=>\(\left(m-1\right)^2=1\)

=>\(\left[{}\begin{matrix}m-1=1\\m-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\\m=0\end{matrix}\right.\)

Vì a*c=-3<0

nên phương trình luôn có 2 nghiệm pb

x1^2+x2^2=10

=>(x1+x2)^2-2x1x2=10

=>(2m+2)^2+6=10

=>(2m+2)^2=4

=>2m+2=2 hoặc 2m+2=-2

=>m=-2 hoặc m=0

PT có 2 nghiệm `<=> \Delta' >0 <=> 2^2-1.(m+1)>0<=> m<3`

Viet: `x_1+x_2=-4`

`x_1 x_2=m+1`

`(x_1)/(x_2)+(x_2)/(x_1)=10/3`

`<=> (x_1^2+x_2^2)/(x_1x_2)=10/3`

`<=> ((x_1+x_2)^2-2x_1x_2)/(x_1x_2)=10/3`

`<=> (4^2-2(m+1))/(m+1)=10/3`

`<=> m=2` (TM)

Vậy `m=2`.

\(x^2+2\left(m+1\right)+4m-4=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=4m-4\end{matrix}\right.\)

Ta có :

\(x_1^2+x_2^2+3x_1x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow\left[-2\left(m+1\right)\right]^2+\left(4m-4\right)=0\)

\(\Leftrightarrow4\left(m^2+2m+1\right)+4m-4=0\)

\(\Leftrightarrow4m^2+8m+4+4m-4=0\)

\(\Leftrightarrow4m^2+12m=0\)

\(\Leftrightarrow4m\left(m+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-3\end{matrix}\right.\)

\(ac=-6< 0\Rightarrow\) phương trình đã cho luôn luôn có 2 nghiệm pb (trái dấu)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=-6\end{matrix}\right.\)

Thế vào đề bài:

\(m-2-3\left(-6\right)=0\)

\(\Leftrightarrow m+16=0\Leftrightarrow m=-16\)

\(x^2-\left(m-2\right)x-6=0\left(1\right)\)

\(\Rightarrow\Delta=b^2-4ac=\left[-\left(m-2\right)\right]^2-4.\left(-6\right)\)

\(=m^2-4m+4+24=m^2-4m+28\)

\(=\left(m-2\right)^2+24\)

Thấy \(\left(m-2\right)^2\ge0\)\(\Rightarrow\left(m-2\right)^2+24>0\forall m\)

Vậy phương trình luân có 2 nghiệm phân biệt \(x_1,x_2\)

Áp dụng \(Vi-ét \) ta có :

\(S=x_1+x_2=\dfrac{-b}{a}=m-2\)

\(P=x_1.x_2=\dfrac{c}{a}=-6\)

Ta có \(x_1+x_2-3.x_1.x_2=0\)

\(\Leftrightarrow m-2-3.\left(-6\right)=0\Rightarrow m=-16\)