\(\Delta=2^2-4\left(m-5\right)\)
\(=4-4m+20=-4m+24\)
Để phương trình có 2 nghiệm thì \(\Delta>=0\)
=>-4m+24>=0
=>-4m>=-24
=>m<=6
Theo Vi-et, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2}{1}=-2\\x_1\cdot x_2=\dfrac{c}{a}=m-5\end{matrix}\right.\)
\(x_2^2-2x_1+m^2-11m+26=0\)
=>\(x_2^2+x_1\left(x_1+x_2\right)+m^2-11m+26=0\)
=>\(\left(x_1^2+x_2^2\right)+x_1x_2+m^2-11m+26=0\)
=>\(\left(x_1+x_2\right)^2-x_1x_2+m^2-11m+26=0\)
=>\(\left(-2\right)^2-\left(m-5\right)+m^2-11m+26=0\)
=>\(4-m+5+m^2-11m+26=0\)
=>\(m^2-12m+35=0\)
=>(m-5)(m-7)=0
=>\(\left[{}\begin{matrix}m-5=0\\m-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=5\left(nhận\right)\\m=7\left(loại\right)\end{matrix}\right.\)
