1. \(2x^2-\left(3m+1\right)x+m^2-m-6=0\)

\(\Delta=b^2-4ac=\left[-\left(3m+1\right)\right]^2-4.2.\left(m^2-m-6\right)=9m^2+6m+1-8m^2+8m+48=m^2+14m+49=\left(m+7\right)^2\ge0\forall m\)

=> PT có 2 nghiệm với mọi m.

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-\left(3m+1\right)\right]}{2}=\dfrac{3m+1}{2}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{m^2-m-6}{2}\end{matrix}\right.\)

Để pt có 2 nghiệm trái dấu \(\Leftrightarrow P< 0\)

\(\Rightarrow\dfrac{m^2-m-6}{2}< 0\Leftrightarrow m^2-m-6< 0\Leftrightarrow-2< m< 3\)

Vậy -2

2. \(mx^2+2\left(m-4\right)x+m+7=0\)

\(\Delta=b^2-4ac=\left[2\left(m-4\right)\right]^2-4.m.\left(m+7\right)=4\left(m^2-8m+16\right)-4m^2-28m=4m^2-32m+64-4m^2-28m=-60m+64\)

Để pt có 2 nghiệm \(\Leftrightarrow\Delta\ge0\)

\(\Rightarrow-60m+64\ge0\Leftrightarrow m\le\dfrac{16}{15}\)

=> PT có 2 nghiệm với \(m\le\dfrac{16}{15}\)

Theo Vi-ét, ta có:

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m-4\right)}{m}=\dfrac{-2m+8}{m}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{m+7}{m}\end{matrix}\right.\)

Ta có hpt: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2m+8}{m}\\x_1-2x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2m+8}{m}\\x_1=2x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\left(2x_2+x_2\right)=-2m+8\\x_1=2x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3mx_2=-2m+8\\x_1=2x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-2m+8}{3m}\\x_1=2.\dfrac{-2m+8}{3m}\end{matrix}\right.\)

Thay \(x_1;x_2\) vào P:

\(\dfrac{2\left(-2m+8\right)}{3m}.\dfrac{-2m+8}{3m}=\dfrac{m+7}{m}\Leftrightarrow\dfrac{2\left(8-2m\right)^2}{9m^2}-\dfrac{m+7}{m}=0\Leftrightarrow\dfrac{2\left(64-32m+4m^2\right)}{9m^2}-\dfrac{9m\left(m+7\right)}{9m^2}=0\Leftrightarrow\dfrac{128-64m+8m^2-9m^2-63m}{9m^2}=0\Leftrightarrow-m^2-127m+128=0\)(1)

Ta có: a+b+c=-1-127+128=0

=> PT (1) có 2 nghiệm \(m_1=1\left(nhận\right);m_2=\dfrac{c}{a}=\dfrac{128}{-1}=-128\left(nhận\right)\)

Vậy m=1;m=-128 thì pt đề cho có 2 nghiệm thỏa đề bài.

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

\(\Delta=b^2-4ac=\left(4m+1\right)^2-4.1.\left[2\left(m-4\right)\right]=16m^2+8m+1-8m+32=16m^2+33>0\forall m\) => PT luôn có 2 nghiệm phân biệt với mọi m.

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(4m+1\right)}{1}=-4m-1\\P=x_1x_2=\dfrac{c}{a}=\dfrac{2\left(m-4\right)}{1}=2m-8\end{matrix}\right.\)

Ta có hpt: \(\left\{{}\begin{matrix}x_1+x_2=-4m-1\\x_1x_2=2m-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=-4m-1\\2x_1x_2=4m-16\end{matrix}\right.\Leftrightarrow x_1+x_2+2x_1x_2=-17\)