a. Em tự giải
b.
\(\Delta'=\left(m-1\right)^2-\left(-m-3\right)=m^2-m+4=\left(m-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0;\forall m\)
Pt luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m-3\end{matrix}\right.\)
\(\left(x_1-x_2\right)^2=4m^2-5x_1+x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4m^2-6x_1+x_1+x_2\)
\(\Leftrightarrow4\left(m-1\right)^2-4\left(-m-3\right)=4m^2-6x_1+2\left(m-1\right)\)
\(\Leftrightarrow6x_1=6m-18\)
\(\Leftrightarrow x_1=m-3\)
Thế vào \(x_1+x_2=2\left(m-1\right)\Rightarrow x_2=2\left(m-1\right)-\left(m-3\right)=m+1\)
Thế vào \(x_1x_2=-m-3\)
\(\Rightarrow\left(m-3\right)\left(m+1\right)=-m-3\)
\(\Leftrightarrow m^2-m=0\)
\(\Rightarrow\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)
