a) Pt: \(x^2-\left(3m-1\right)x+2m^2-m=0\left(1\right)\)
Với m = 1 ta có pt:
\(x^2-\left(3.1-1\right)x+2.1^2-1=0\)
\(x^2-2x+1=0\)
Ta có: \(\Delta=b^2-4ac=\left(-2\right)^2-4.1.1=0\)
Nên pt có nghiệm kép: \(x_1=x_2=\frac{-b}{a}=\frac{-\left(-2\right)}{2}=1\)
b) +) Pt (1) có 2 nghiệp phân biệt \(\Leftrightarrow\Delta>0\)
\(\Leftrightarrow\left[-\left(3m-1\right)^2\right]-4.1\left(2m^2-3\right)>0\)
\(\Leftrightarrow\left(3m-1\right)^2-8m^2+4m>0\)
\(\Leftrightarrow9m^2-6m+1-8m^2+4m>0\)
\(\Leftrightarrow m^2-2m+1>0\)
\(\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m-1\ne0\Leftrightarrow m\ne1\)
+) \(\left|x_1-x_2\right|-2=0\)
\(\Leftrightarrow\left|x_1-x_2\right|=2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4\left(2\right)\)
Áp dụng hệ thức Vi- ét cho pt(1):
\(\left\{{}\begin{matrix}x_1+x_2=3m-1\\x_1.x_2=2m^2-m\end{matrix}\right.\)
Thay vào pt (2) ta được:
\(\Leftrightarrow\left(3m-1\right)^2-4.\left(2m^2-m\right)=4\)
\(\Leftrightarrow9m^2-6m+1-8m^2+4m=4\)
\(\Leftrightarrow m^2-2m+1=4\)
\(\Leftrightarrow m^2-2m-3=0\)
\(\Leftrightarrow\left(m+1\right)\left(m-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m+1=0\\m-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-1\\m=3\end{matrix}\right.\)(tmđk)
Vậy ..............
