\(\Delta=\left(2m+2\right)^2-4\cdot2\cdot\left(m^2+4m+3\right)\)
\(=4m^2+8m+4-8m^2-32m-24=-4m^2-24m-20\)
\(=-4\left(m^2+6m+5\right)=-4\left(m+1\right)\left(m+5\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>-4(m+1)(m+5)>=0
=>(m+1)(m+5)<=0
=>-5<=m<=-1
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=\frac{-2\left(m+1\right)}{2}=-\left(m+1\right)\\ x_1x_2=\frac{c}{a}=\frac{m^2+4m+3}{2}\end{cases}\)
\(A=\left|x_1x_2-2x_1-2\cdot x_2\right|\)
\(=\left|\frac{m^2+4m+3}{2}-2\left(x_1+x_2\right)\right|=\left|\frac{m^2+4m+3}{2}-2\cdot\left(-m-1\right)\right|\)
\(=\left|\frac{m^2+4m+3}{2}+2\left(m+1\right)\right|=\left|\frac{m^2+4m+3+4\left(m+1\right)}{2}\right|\)
\(=\left|\frac{m^2+8m+7}{2}\right|\)
\(=\frac12\cdot\left|\left(m+1\right)\left(m+7_{}\right)\right|\)
-5<=m<=-1
=>m+1<=0 và m+5>=0
=>m+1<=0 và m+5+2>=2
=>m+1<=0 và m+7>0
=>(m+1)(m+7)<=0
=>\(A=-\frac12\left(m+1\right)\left(m+7\right)\)
\(=-\frac12\left\lbrack\left(m+4-3\right)\left(m+4+3\right)\right\rbrack=-\frac12\cdot\left\lbrack\left(m+4\right)^2-9\right\rbrack=-\frac12\left(m+4\right)^2+\frac92\le\frac92\forall x\)
Dấu '=' xảy ra khi m+4=0
=>m=-4
