TH1: m=1
Phương trình sẽ trở thành:
\(\left(1-1\right)x^2-2\left(1-4\right)x+1-5=0\)
=>6x-4=0
=>6x=4
=>\(x=\dfrac{2}{3}\)
=>Phương trình có 1 nghiệm duy nhất
=>Loại
TH2: m<>1
\(\text{Δ}=\left[-2\left(m-4\right)\right]^2-4\left(m-1\right)\left(m-5\right)\)
\(=4\left(m^2-8m+16\right)-4\left(m^2-6m+5\right)\)
\(=4\left(m^2-8m+16-m^2+6m-5\right)\)
\(=4\left(-2m+11\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>-2m+11>=0
=>-2m>=-11
=>\(m< =\dfrac{11}{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{2\left(m-4\right)}{m-1}\\x_1x_2=\dfrac{c}{a}=\dfrac{m-5}{m-1}\end{matrix}\right.\)
\(x_1^2+x_2^2=5\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=5\)
=>\(\dfrac{4\left(m-4\right)^2}{\left(m-1\right)^2}-\dfrac{2\left(m-5\right)}{m-1}-5=0\)
=>\(\dfrac{4\left(m-4\right)^2-2\left(m-5\right)\left(m-1\right)}{\left(m-1\right)^2}-5=0\)
=>\(4\left(m^2-8m+16\right)-2\left(m^2-6m+5\right)-5\left(m-1\right)^2=0\)
=>\(4m^2-32m+64-2m^2+12m-10-5m^2+10m-5=0\)
=>\(-3m^2-10m+49=0\)
=>\(\left[{}\begin{matrix}m=\dfrac{-5+2\sqrt{43}}{3}\left(nhận\right)\\m=\dfrac{-5-2\sqrt{43}}{3}\left(nhận\right)\end{matrix}\right.\)
