a: \(\text{Δ}=\left(2m-1\right)^2-8\left(m-1\right)\)
\(=4m^2-4m+1-8m+8\)
\(=4m^2-12m+9=\left(2m-3\right)^2>=0\)
=>Pt luôn có hai nghiệm phân biệt
Theo đề, ta có: \(\left\{{}\begin{matrix}3x_1-4x_2=11\\x_1+x_2=\dfrac{-2m+1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x_1-4x_2=11\\3x_1+3x_2=\dfrac{-6m+3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7\cdot x_2=11-\dfrac{-6m+3}{2}=\dfrac{22+6m-3}{2}=\dfrac{6m+19}{2}\\3x_1=11+4x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-6m-19}{14}\\x_1=\dfrac{1}{3}\left(4\cdot\dfrac{-6m-19}{14}+11\right)=\dfrac{1}{3}\left(\dfrac{-12m-38}{7}+11\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-6m-19}{14}\\x_1=\dfrac{1}{3}\cdot\dfrac{-12m-38+77}{7}=\dfrac{-4m+13}{7}\end{matrix}\right.\)
Theo đề, ta có: \(\dfrac{-6m-19}{14}\cdot\dfrac{-4m+13}{7}=\dfrac{m-1}{2}\)
\(\Leftrightarrow m=-2\)
b: Để phươg trình có hai nghiệm đều dương thì (-2m+1)/2>0 và (m-1)/2>0
=>-2m+1>0 và m-1>0
=>m<1/2 và m>1
hay \(m\in\varnothing\)
