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:

a: Thay m=-5 vào (1), ta được:

\(x^2+2\left(-5+1\right)x-5-4=0\)

\(\Leftrightarrow x^2-8x-9=0\)

=>(x-9)(x+1)=0

=>x=9 hoặc x=-1

b: \(\text{Δ}=\left(2m+2\right)^2-4\left(m-4\right)=4m^2+8m+4-4m+16=4m^2+4m+20>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt 

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\)

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

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

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

\(\Leftrightarrow4m^2+9m=0\)

=>m(4m+9)=0

=>m=0 hoặc m=-9/4

Δ=(m+2)^2-4*2m=(m-2)^2

Để PT có hai nghiệm pb thì m-2<>0

=>m<>2

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{m+2}{2m}=\dfrac{2m}{4}=\dfrac{m}{2}\)

=>2m^2=2m+4

=>m^2-m-2=0

=>m=2(loại) hoặc m=-1

Để pt có 2 nghiệm trái dấu \(\Leftrightarrow3m-2< 0\Leftrightarrow m< \dfrac{2}{3}\)

Nếu \(x_1< 0\) thì \(\dfrac{1}{x_1}-3< 0\) trong khi \(\left|\dfrac{1}{x_2}\right|>0\Rightarrow\) không thỏa mãn

Vậy \(x_1>0;x_2< 0\)

Do đó:

\(\dfrac{1}{x_1}-3=\left|\dfrac{1}{x_2}\right|=-\dfrac{1}{x_2}\)

\(\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}=3\Leftrightarrow x_1+x_2-3x_1x_2=0\)

\(\Leftrightarrow-2\left(m-1\right)-3\left(3m-2\right)=0\)

\(\Leftrightarrow m=...\)

\(\Delta'=\left(m+1\right)^2-5\ge0\Leftrightarrow m^2+2m-4\ge0\) (1)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=5\end{matrix}\right.\)

\(\dfrac{1}{\left|x_1\right|}+\dfrac{1}{\left|x_2\right|}=2\Leftrightarrow\dfrac{\left|x_1\right|+\left|x_2\right|}{\left|x_1x_2\right|}=2\)

\(\Leftrightarrow\left|x_1\right|+\left|x_2\right|=2\left|x_1x_2\right|=10\)

\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=100\)

\(\Leftrightarrow x_1^2+x_2^2+10=100\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=90\)

\(\Leftrightarrow4\left(m+1\right)^2-10=90\)

\(\Leftrightarrow\left(m+1\right)^2=25\Rightarrow\left[{}\begin{matrix}m=4\\m=-6\end{matrix}\right.\) 

Thế vào (1) kiểm tra thấy đều thỏa mãn, vậy...

pt sai 

Để pt có 2 nghiệm pb khác 0:

\(\left\{{}\begin{matrix}\Delta'=4\left(m-1\right)^2-3\left(m^2-4m+1\right)>0\\x_1x_2=\dfrac{m^2-4m+1}{3}\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2+4m+1>0\\m^2-4m+1\ne0\end{matrix}\right.\) (1)

Theo hệ thức Viet:  \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{4\left(m-1\right)}{3}\\x_1x_2=\dfrac{m^2-4m+1}{3}\end{matrix}\right.\)

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\left(x_1+x_2\right)\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{2}\left(x_1+x_2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1+x_2=0\\x_1x_2=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4\left(m-1\right)}{3}=0\\\dfrac{m^2-4m+1}{3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\\m=5\end{matrix}\right.\) 

Thế vào hệ điều kiện (1) kiểm tra chỉ có \(\left[{}\begin{matrix}m=1\\m=5\end{matrix}\right.\) thỏa mãn