XétΔ′=(m+1)2−2.(m2+4m+3)=−m2−6m−5≥0⇔−5≤m≤−1.Khi đó, theo định lí Viet:$x1+x2=\frac{-2.(m+1)}{2}=-(m+1)$$x1.x2=\frac{m^2+4m+3}{2}$Thay vào ta có:$A= |x1.x2-2.(x1+x2)|=|\frac{m^2+4m+3}{2}+2m+2|=|\frac{m^2+8m+7}{2}|$$A=|\frac{(m^2+8m+16)-9}{2}|=\frac{|(m+4)^2-9|}{2}$$\rightarrow$ $A$ max khi $m=-4VậyAmax =\frac{9}{2} $
XétΔ′=(m+1)2−2.(m2+4m+3)=−m2−6m−5≥0⇔−5≤m≤−1.Khi đó, theo định lí Viet:x1+x2=2.(m+1)2=(m+1)x1.x2=m2+4m+32Thay vào ta có:$A= |x1.x2-2.(x1+x2)|=|\frac{m^2+4m+3}{2}-2m-2|=|\frac{m^2-1}{2}|$mà $-5\leq m\leq -1$$\rightarrow$ $A$ max khi $m=-5VậyAmax =12$