\(\Delta=\left(3m-2\right)^2+12\left(3m+1\right)=9m^2+24m+16=\left(3m+4\right)^2\)
\(\Rightarrow\) Pt luôn có nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3m-2}{3}\\x_1x_2=-\dfrac{3m+1}{3}\end{matrix}\right.\)
a.
Pt có 2 nghiệm thỏa mãn \(-2< x_1\le x_2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1+2\right)\left(x_2+2\right)>0\\\dfrac{x_1+x_2}{2}>-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2+2\left(x_1+x_2\right)+4>0\\x_1+x_2>-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3m+1}{3}+\dfrac{2\left(3m-2\right)}{3}+4>0\\\dfrac{3m-2}{3}>-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3m+7>0\\3m>-10\end{matrix}\right.\) \(\Rightarrow m>-\dfrac{7}{3}\)
b.
\(2x_1^2-9x_2^2+7x_1x_2=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(2x_1+9x_2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=x_2\\x_1=-\dfrac{9}{2}x_2\end{matrix}\right.\)
TH1: \(x_1=x_2\Rightarrow\Delta=0\)
\(\Rightarrow3m+4=0\Rightarrow m=-\dfrac{4}{3}\)
TH2: \(x_1=-\dfrac{9}{2}x_2\)
Thay vào \(x_1+x_2=\dfrac{3m-2}{3}\)
\(\Rightarrow-\dfrac{9}{2}x_2+x_2=\dfrac{3m-2}{3}\Rightarrow x_2=\dfrac{-2\left(3m-2\right)}{21}\)
\(\Rightarrow x_1=-\dfrac{9}{2}x_2=\dfrac{3\left(3m-2\right)}{7}\)
Thay vào \(x_1x_2=-\dfrac{3m+1}{3}\)
\(\Rightarrow\dfrac{-2\left(3m-2\right)}{21}.\dfrac{3\left(3m-2\right)}{7}=-\dfrac{3m+1}{3}\)
\(\Leftrightarrow6\left(3m-2\right)^2-49\left(3m+1\right)=0\)
\(\Leftrightarrow54m^2-219m-25=0\)
\(\Rightarrow\left[{}\begin{matrix}m=-\dfrac{1}{9}\\m=\dfrac{25}{6}\end{matrix}\right.\)
c.
Đặt \(f\left(x\right)=3x^2-\left(3m-2\right)x-\left(3m+1\right)\)
Để \(x_1< 3< x_2\)
\(\Rightarrow f\left(3\right)< 0\)
\(\Rightarrow3.3^2-3\left(3m-2\right)-\left(3m-1\right)< 0\)
\(\Rightarrow32-12m< 0\)
\(\Rightarrow m>\dfrac{8}{3}\)
