Question

Answer 1

a: Thay m=2 vào (1), ta được:

\(x^2+2x+2-1=0\)

=>\(x^2+2x+1=0\)

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1

b: \(\text{Δ}=m^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>=0\forall m\)

=>Phương trình luôn có nghiệm với mọi m

c: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-m\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(A=\left(x_1+1\right)^2\cdot\left(x_2+1\right)^2+2016\)

\(=\left[\left(x_1+1\right)\left(x_2+1\right)\right]^2+2016\)

\(=\left[x_1x_2+x_1+x_2+1\right]^2+2016\)

\(=\left[m-1-m+1\right]^2+2016=2016\)