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

   \(=4m^2+8m+4-4m^2-8\)

   \(=8m-4\)

Để pt có 2 nghiệm thì \(\Delta>0\)

                                    \(\Leftrightarrow8m-4>0\)

                                      \(\Leftrightarrow m>\dfrac{1}{2}\)

Theo hệ thức Vi-ét, ta có:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)

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

\(\Leftrightarrow x_1^2+m^2+2+2=2x_1+2\left(m+1\right)\)

\(\Leftrightarrow x_1^2-2x_1+4+m^2-2m-2=0\)

\(\Leftrightarrow x_1^2-2x_1+2+m^2-2m=0\)

\(\Leftrightarrow x_1^2-2x_1+1+m^2-2m+1=0\)

\(\Leftrightarrow\left(x_1-1\right)^2+\left(m-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=1\\m=1\end{matrix}\right.\)(tm)

Vậy \(m=1\)

Δ=(m+2)^2-4*2m

=m^2+4m+4-8m

=(m-2)^2>=0

Để PT luôn có hai nghiệm phân biệt thì m-2<>0

=>m<>2

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

=>(m+2)^2-2m<=3

=>m^2+4m+4-2m-3<=0

=>m^2+2m+1<=0

=>(m+1)^2<=0

=>m=-1

△'=(-2)²-1(m-1)

   =4-m+1

   =5-m

Để PT có 2 no pb thì △'>0

⇒5-m>0

⇒m<5

theo vi-ét ta có

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

mà: \(x^2_1x_2+x_1x_2^2-2\left(x_1+x_2\right)=0\)

⇔\(\left(x_1x_2\right)\left(x_1+x_2\right)-2\left(x_1+x_2\right)=0\)

⇔\(\left(m-1\right)4-2\cdot4=0\)

⇔\(4m-4-8=0\)

⇔4m-12=0

⇔4m=12

⇔m=3

Vậy ...

b) phương trình có 2 nghiệm  \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)

\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)

\(\Leftrightarrow-4m+4\ge0\)

\(\Leftrightarrow m\le1\)

Ta có: \(x_1^2+x_1x_2+x_2^2=1\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

x2^2-x1x2+2(m-2)x1=m^2-6m+23

=>x2^2+x1(x1+x2)-x1x2=m^2-6m+23

=>(x1+x2)^2-2x1x2=m^2-6m+23

=>(2m-4)^2-2(-7)=m^2-6m+23

=>4m^2-16m+16+14-m^2+6m-23=0

=>m=7/3 hoặc m=1

\(\Delta'=\left(m-1\right)^2+m^3-\left(m+1\right)^2=m^3-4m\ge0\) \(\Rightarrow\left[{}\begin{matrix}m\ge2\\-2\le m\le0\end{matrix}\right.\)

Theo hệ thức Viet:  \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m^3+\left(m+1\right)^2\end{matrix}\right.\)

Do \(x_1+x_2\le4\Rightarrow m-1\le2\Rightarrow m\le3\)

\(\Rightarrow\left[{}\begin{matrix}2\le m\le3\\-2\le m\le0\end{matrix}\right.\)

\(P=x_1^3+x_2^3+3x_1x_2\left(x_1+x_2\right)+8x_1x_2\)

\(=\left(x_1+x_2\right)^3+8x_1x_2\)

\(=8\left(m-1\right)^3+8\left[-m^3+\left(m+1\right)^2\right]\)

\(=8\left(5m-2m^2\right)\)

\(P=8\left(5m-2m^2-2+2\right)=16-8\left(m-2\right)\left(2m-1\right)\le16\)

\(P_{max}=16\) khi \(m=2\)

\(P=8\left(5m-2m^2+18-18\right)=8\left(9-2m\right)\left(m+2\right)-144\ge-144\)

\(P_{min}=-144\) khi \(m=-2\)

PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta=\left(2m-3\right)^2-4\left(m-3\right)=9>0\)

Vậy PT có 2 nghiệm phân biệt với mọi m

Ta có \(\left[{}\begin{matrix}x_1=\dfrac{2m-3+3}{2}=m\\x_2=\dfrac{2m-3-3}{2}=m-3\end{matrix}\right.\)

Ta thấy \(m>m-3\) nên \(1< m-3< m< 6\Leftrightarrow4< m< 6\)

Vậy \(4< m< 6\)  thỏa yêu cầu đề