a) Khi \(m=0\)
\(x^2-\left(5m-1\right)x+6m^2-2m\)
\(\Leftrightarrow x^2+x\)
\(\Delta=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=0\end{matrix}\right.\)
Vậy khi m = 0 thì pt có 2no \(x_1=-1;x_2=0\)
b) \(\Delta=\left(-\left(5m-1\right)\right)^2-4\left(6m^2-2m\right)\)
\(=\left(5m-1\right)^2-4\left(6m^2-2m\right)\)
\(=25m^2-10m+1-24m^2+8m\)
\(=m^2-2m+1\)
\(=\left(m-1\right)^2\ge0\forall m\)
Vì \(\Delta\ge0\) nên pt có 2no phân biệt thỏa mãn hệ thức Vi-ét:
\(\left\{{}\begin{matrix}x_1+x_2=5m-1\\x_1x_2=6m^2-2m\end{matrix}\right.\)
Để \(x_1+x_2=1\)
\(\Leftrightarrow5m-1=1\)
\(\Leftrightarrow5m=2\)
\(\Leftrightarrow m=\frac{2}{5}\)
