a: Thay m=-1 vào (1), ta được:
\(x^2+2x-\left(-1\right)=0\)
=>\(x^2+2x+1=0\)
=>\(\left(x+1\right)^2=0\)
=>x+1=0
=>x=-1
b: \(\text{Δ}=2^2-4\cdot1\cdot\left(-m\right)=4m+4\)
Để phương trình (1) có nghiệm thì Δ>=0
=>4m+4>=0
=>m>=-1
c: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1x_2=\dfrac{c}{a}=-m\end{matrix}\right.\)
\(P=x_1^4+x_2^4\)
\(=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2\)
\(=\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2-2\left(x_1x_2\right)^2\)
\(=\left[4-2\cdot\left(-m\right)\right]^2-2\cdot\left(-m\right)^2\)
\(=\left(2m+4\right)^2-2m^2\)
\(=4m^2+16m+16-2m^2\)
\(=2m^2+16m+16\)
\(=2\left(m^2+8m+8\right)\)
\(=2\left(m^2+8m+16-8\right)\)
\(=2\left(m+4\right)^2-16>=-16\forall m\)
Dấu '=' xảy ra khi m=-4
