Bài 2.
a.\(x=2\)
\(\Leftrightarrow2^2-6.2+m+1=0\Leftrightarrow m+1=8\Leftrightarrow m=7\)
b.Theo hệ thức Vi-ét, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=6\\x_1.x_2=m+1\end{matrix}\right.\)
\(x_1^2+x_2^2=26\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1.x_2=26\)
\(\Leftrightarrow6^3-2\left(m+1\right)=26\)
\(\Leftrightarrow-2m-1=-10\)
\(\Leftrightarrow m=\dfrac{9}{2}\)