\(\Delta'=\left(m-3\right)^2+m+1=m^2-6m+9+m+1=m^2-5m+10=m^2-2.m\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}+10=\left(m-\dfrac{5}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\forall m\)
=> Pt luôn có 2 no phân biệt
Theo định lý viet\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-3\right)\\x_1x_2=-m-1\end{matrix}\right.\)
Ta có\(x_1^2+x_2^2=10\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)\(\Leftrightarrow2\left(m-3\right)^2+m+1=5\)\(\Leftrightarrow2m^2-12m+18+m+1-5=0\)\(\Leftrightarrow2m^2-11m+14=0\)\(\Leftrightarrow\left(m-2\right)\left(2m-7\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}m=2\\m=\dfrac{7}{2}\end{matrix}\right.\)
\(\Delta'=\left(m-3\right)^2+\left(m+1\right)=m^2-5m+10=\left(m-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0;\forall m\)
\(\Rightarrow\)Phương trình đã cho luôn có 2 nghiệm pb với mọi m
b. Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-3\right)\\x_1x_2=-m-1\end{matrix}\right.\)
\(x_1^2+x_2^2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)
\(\Leftrightarrow4\left(m-3\right)^2+2\left(m+1\right)=10\)
\(\Leftrightarrow4m^2-22m+28=0\Rightarrow\left[{}\begin{matrix}m=2\\m=\dfrac{7}{2}\end{matrix}\right.\)
