a: \(\text{Δ}=\left[2\left(m+1\right)\right]^2-4\cdot1\left(m^2-2m-5\right)\)
\(=4m^2+8m+4-4m^2+8m+20\)
=16m+24
Để phương trình (1) có hai nghiệm thì Δ>=0
=>16m+24>=0
=>16m>=-24
=>\(m>=-\dfrac{24}{16}=-\dfrac{3}{2}\)
b: Theo Vi-et, ta có: \(x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right);x_1x_2=\dfrac{c}{a}=m^2-2m-5\)
\(3x_1+3x_2=-\dfrac{1}{2}x_1x_2\)
=>\(3\left(x_1+x_2\right)=-\dfrac{1}{2}x_1x_2\)
=>\(3\cdot2\left(m+1\right)=-\dfrac{1}{2}\left(m^2-2m-5\right)\)
=>\(6\left(m+1\right)+\dfrac{1}{2}m^2-m-\dfrac{5}{2}=0\)
=>\(\dfrac{1}{2}m^2-5m+\dfrac{7}{2}=0\)
=>\(m^2-10m+7=0\)
=>(m-2)(m-5)=0
=>\(\left[{}\begin{matrix}m=2\left(nhận\right)\\m=5\left(nhận\right)\end{matrix}\right.\)
