a) Ta có: \(\text{Δ}=\left(2m\right)^2-4\cdot1\cdot\left(-3m-2\right)=4m^2+12m+8=4m^2+12m+9-1=\left(2m+3\right)^2-1\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
\(\Leftrightarrow\left(2m+3\right)^2>1\)
\(\Leftrightarrow\left[{}\begin{matrix}2m+3>1\\2m+3< -1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2m>-2\\2m< -4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>-1\\m< -2\end{matrix}\right.\)
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=-2m\\x_1\cdot x_2=-3m-2\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2m\\2x_1-3x_2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x_1+2x_2=-4m\\2x_1-3x_2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x_2=-4m-1\\x_1+x_2=-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-4m-1}{5}\\x_1=-2m+\dfrac{4m+1}{5}=\dfrac{-6m+1}{5}\end{matrix}\right.\)
Ta có: \(x_1\cdot x_2=-3m-2\)
\(\Leftrightarrow\dfrac{-4m-1}{5}\cdot\dfrac{-6m+1}{5}=-3m-2\)
\(\Leftrightarrow\left(-4m-1\right)\left(-6m+1\right)=25\left(-3m-2\right)\)
\(\Leftrightarrow24m^2-4m+6m-1=-75m+50\)
\(\Leftrightarrow24m^2+2m-1+75m-50=0\)
\(\Leftrightarrow24m^2+77m-51=0\)
Đến đây bạn tự làm nhé