\(\left(x-m+1\right)\left(x+2m-3\right)=0\)
\(\Leftrightarrow x^2+2xm-3x-xm-2m^2+3m+x+2m-3=0\)
\(\Leftrightarrow-2m^2+m\left(x+5\right)+\left(x^2-2x-3\right)=0\)
Để pt có 2 nghiệm phân biệt thì \(\Delta>0\)
\(\Leftrightarrow\left(x+5\right)^2-4\left(x^2-2x-3\right)\left(-2\right)>0\)
\(\Leftrightarrow x^2+10x+25+8\left(x^2-2x-3\right)>0\)
\(\Leftrightarrow x^2+10x+25+8x^2-16x-24>0\)
\(\Leftrightarrow9x^2-6x+1>0\)
\(\Leftrightarrow\left(3x-1\right)^2>0\)
\(\Leftrightarrow\left|3x-1\right|>0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1< 0\\3x-1>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \frac{1}{3}\\x>\frac{1}{3}\end{matrix}\right.\)
Vậy....
