- Xét phương trình có : \(\left\{{}\begin{matrix}a=1\\b=-2\left(m-1\right)=2-2m\\c=-4m\end{matrix}\right.\)
=> \(\Delta=b^2-4ac=\left(2-2m\right)^2-4.1.\left(-4m\right)\)
=> \(\Delta=4-8m+4m^2+16m\)
=> \(\Delta=4m^2+8m+4\)
=> \(\Delta=\left(2m+2\right)^2\ge0\)
- Để phương trình có 2 nghiệm phân biệt thì : \(2m+2\ne0\)
=> \(m\ne-1\)
- Nên phương trình có 2 nghiệm phân biệt là :
\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-2\left(m-1\right)\right)-\sqrt{\left(2m+2\right)^2}}{2.1}=\frac{2m-2-\left|2m+2\right|}{2}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-2\left(m-1\right)\right)+\sqrt{\left(2m+2\right)^2}}{2a}=\frac{2m-2+\left|2m+2\right|}{2}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x_1=\frac{2\left(m-1-\left|m+1\right|\right)}{2}=m-1-\left|m+1\right|\\x_2=\frac{2\left(m-1+\left|m+1\right|\right)}{2}=m-1+\left|m+1\right|\end{matrix}\right.\)
Ta có : \(3x_1-x_2=2\)
TH1 : \(\left\{{}\begin{matrix}x_1=m-1-\left|m+1\right|\\x_2=m-1+\left|m+1\right|\end{matrix}\right.\)
=> \(3\left(m-1-\left|m+1\right|\right)-\left(m-1+\left|m+1\right|\right)=2\)
=> \(3m-3-3\left|m+1\right|-m+1-\left|m+1\right|=2\)
=> \(2m-4-4\left|m+1\right|=0\)
=> \(\left|m+1\right|=\frac{2m-4}{4}\)
=> \(\left[{}\begin{matrix}m+1=\frac{2m-4}{4}\\m+1=\frac{4-2m}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}4m+4=2m-4\\4m+4=4-2m\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2m=-8\\6m=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}m=-4\\m=0\end{matrix}\right.\) ( TM )
TH2 : \(\left\{{}\begin{matrix}x_1=m-1+\left|m+1\right|\\x_2=m-1-\left|m+1\right|\end{matrix}\right.\)
=> \(3\left(m-1+\left|m+1\right|\right)-\left(m-1-\left|m+1\right|\right)=2\)
=> \(3m-3+3\left|m+1\right|-m+1+\left|m+1\right|=2\)
=> \(4\left|m+1\right|-4+2m=0\)
=> \(\left|m+1\right|=\frac{4-2m}{4}\)
=> \(\left[{}\begin{matrix}m+1=\frac{2m-4}{4}\\m+1=\frac{4-2m}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}4m+4=2m-4\\4m+4=4-2m\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2m=-8\\6m=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}m=-4\\m=0\end{matrix}\right.\) ( TM )
Vậy m = 0 và m = -4 thỏa mãn điều kiện trên .
