a: Thay m=1 vào (1), ta được:
\(x^2-\left(1+5\right)x-1+6=0\)
=>\(x^2-6x+5=0\)
=>(x-1)(x-5)=0
=>\(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b: Thay x=-2 vào (1), ta được:
\(\left(-2\right)^2-\left(-2\right)\left(m+5\right)-m+6=0\)
=>\(4+2m+10-m+6=0\)
=>m+20=0
=>m=-20
c: \(\text{Δ}=\left[\left(m+5\right)\right]^2-4\left(-m+6\right)\)
\(=m^2+10m+25+4m-24\)
\(=m^2+14m+1\)
Để phương trình (1) có nghiệm thì Δ>=0
=>\(m^2+14m+1>=0\)
=>\(\left(m+7\right)^2-48>=0\)
=>\(\left(m+7\right)^2>=48\)
=>\(\left[{}\begin{matrix}m+7>=4\sqrt{3}\\m+7< =-4\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>=4\sqrt{3}-7\\m< =-4\sqrt{3}-7\end{matrix}\right.\)
Theo Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+5\\x_1x_2=\dfrac{c}{a}=-m+6\end{matrix}\right.\)
\(x_1^2\cdot x_2+x_1\cdot x_2^2=24\)
=>\(x_1x_2\left(x_1+x_2\right)=24\)
=>\(\left(m+5\right)\left(-m+6\right)=24\)
=>\(-m^2+6m-5m+30-24=0\)
=>\(-m^2+m+6=0\)
=>\(m^2-m-6=0\)
=>(m-3)(m+2)=0
=>\(\left[{}\begin{matrix}m=3\left(nhận\right)\\m=-2\left(loại\right)\end{matrix}\right.\)
