Phương trình có hai nghiệm phân biệt khi:\(\Delta'=\left(-m\right)^2-1\left(2m-1\right)>0\)\(\Leftrightarrow m^2-2m+1>0\Leftrightarrow\left(m-1\right)^2>0\)\(\Rightarrow m\ne1.\)Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=2m-1\end{matrix}\right.\).Từ đề bài, suy ra: \(\left[x_1^2-x_1\left(x_1+x_2\right)+3\right]\left[x_2^2-\left(x_1+x_2\right)x_2-2\right]=50\)\(\Leftrightarrow\left(3-x_1x_2\right)\left(-x_1x_2-2\right)=50\)\(\Leftrightarrow\left(3-2m+1\right)\left(1-2m-2\right)=50\)\(\Leftrightarrow\left(2-m\right)\left(2m+1\right)=-25\)\(\Leftrightarrow-2m^2+3m+27=0\)\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{9}{2}\\m=-3\end{matrix}\right.\left(TM\right)\)Vậy: \(m=\dfrac{9}{2}\) hoặc \(m=-3\)Câu trả lời: Phương trình có hai nghiệm phân biệt khi:\(\Delta'=\left(-m\right)^2-1\left(2m-1\right)>0\)\(\Leftrightarrow m^2-2m+1>0\Leftrightarrow\left(m-1\right)^2>0\)\(\Rightarrow m\ne1.\)Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=2m-1\end{matrix}\right.\).Từ đề bài, suy ra: \(\left[x_1^2-x_1\left(x_1+x_2\right)+3\right]\left[x_2^2-\left(x_1+x_2\right)x_2-2\right]=50\)\(\Leftrightarrow\left(3-x_1x_2\right)\left(-x_1x_2-2\right)=50\)\(\Leftrightarrow\left(3-2m+1\right)\left(1-2m-2\right)=50\)\(\Leftrightarrow\left(2-m\right)\left(2m+1\right)=-25\)\(\Leftrightarrow-2m^2+3m+27=0\)\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{9}{2}\\m=-3\end{matrix}\right.\left(TM\right)\)Vậy: \(m=\dfrac{9}{2}\) hoặc \(m=-3\)