http://olm.vn/hoi-dap/question/91925.html

Ptr có nghiệm `<=>\Delta' >= 0`

                       `<=>[-(m+1)^2]-6m+4 >= 0`

                      `<=>m^2+2m+1-6m+4 >= 0`

                      `<=>m^2-4m+5 >= 0<=>(m-2)^2+1 >= 0` (LĐ `AA m`)

`=>` Áp dụng Viét có: `{(x_1+x_2=[-b]/a=2m+2),(x_1.x_2=c/a=6m-4):}`

Có:`(2m-2)x_1+x_2 ^2-4x_2=4`

`<=>(x_1+x_2-4)x_1+x_2 ^2-4x_2=4`

`<=>x_1 ^2+x_1 x_2 -4x_1+x_2 ^2-4x_2=4`

`<=>(x_1+x_2)^2-x_1x_2-4(x_1+x_2)=4`

`<=>(2m+2)^2-(6m-4)-4(2m+2)=4`

`<=>4m^2+8m+4-6m+4-8m-8=4`

`<=>4m^2-6m-4=0`

`<=>(2m-3/2)^2-25/4=0`

`<=>|2m-3/2|=5/2`

`<=>[(m=2),(m=-1/2):}`

ghi ko hiểu j hết mẹ!!???????

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=2m+1\end{matrix}\right.\)

Pt có 4 nghiệm pb khi: \(\left\{{}\begin{matrix}2m+1>0\\2m+1\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m\ne0\end{matrix}\right.\)

Do \(x=\pm1< 3\) nên để  \(x_1< x_2< x_3< x_4< 3\) thì:

\(\sqrt{2m+1}< 3\Leftrightarrow m< 4\) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}< m< 4\\m\ne0\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}x_1-x_3=x_3-x_2\\x_1-x_3=x_2-x_1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-x_2\\x_1-x_3=-x_1-x_1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_2=-x_1\\x_3=3x_1\end{matrix}\right.\)

Do vai trò \(x_1;x_2\) như nhau, giả sử \(x_1< 0\) \(\Rightarrow x_1;x_3\) là 2 nghiệm âm

TH1: \(\left\{{}\begin{matrix}x_1=-1\\x_2=1\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}x_3=-\sqrt{2m+1}\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-\sqrt{2m+1}=-3\Rightarrow m=4\)

TH2: \(x_1=-\sqrt{2m+1}\Rightarrow\left\{{}\begin{matrix}x_3=-1\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-1=-3\sqrt{2m+1}\) \(\Rightarrow m=-\dfrac{4}{9}\)