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:

Đề nhìn sao cứ sai sai vậy bạn :v

hình như đề thiếu hả bạn

\(\text{Δ}=\left[-\left(m+1\right)\right]^2-4\cdot1\cdot m\)

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

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

=>Phương trình luôn có hai nghiệm

Theo Vi-et, ta có:

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

\(x_1^2+x_2^2=\left(x_1-1\right)\left(x_2-1\right)-x_1-x_2+5\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2=x_1x_2-2\left(x_1+x_2\right)+6\)

=>\(\left(m+1\right)^2-2m=m-2\left(m+1\right)+6\)

=>\(m^2+1=m-2m-2+6\)

=>\(m^2+1=-m+4\)

=>\(m^2+m-3=0\)

=>\(m=\dfrac{-1\pm\sqrt{13}}{2}\)

\(\Delta=1-4m>0\Rightarrow m< \dfrac{1}{4}\)

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

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

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

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

\(\Leftrightarrow m^2-m-1=1\)

\(\Leftrightarrow m^2-m-2=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=2>\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)