a/ Ta có: $a=1,b=-(5m-1),c=6m^2-2m$
$Δ=b^2-4ac\\=[-(5m-1)]^2-4.1.(6m^2-2m)\\=(5m-1)^2-24m^2+8m\\=25m^2-10m+1-24m^2+8m\\=m^2-2m+1\\=(m-1)^2\ge 0$
$→$ Pt có luôn có 2 nghiệm
b/ Theo Vi-ét:
$\begin{cases}x_1+x_2=\dfrac{-b}{a}=5m-1\\x_1x_2=\dfrac{c}{a}=6m^2-2\end{cases}$
$x_1^2+x_2^2=1\\↔x_1^2+2x_1x_2+x_2^2-2x_1x_2=1\\↔(x_1^2+2x_1x_2+x_2^2)-2x_1x_2=1\\↔(x_1+x_2)^2-2x_1x_1=1\\↔(5m-1)^2-2(6m^2-2m)=1\\↔25m^2-10m+1-12m^2+4m=1\\↔13m^2-6m=0\\↔m(13m-6)=0\\↔\left[\begin{array}{1}m=0\\13m-6=0\end{array}\right.\\↔\left[\begin{array}{1}m=0\\13m=6\end{array}\right.\\↔\left[\begin{array}{1}m=0\\m=\dfrac{6}{13}\end{array}\right.$
Vậy $m∈\left\{0;\dfrac{6}{13}\right\}$
