\(\Delta'=\left(m+4\right)^2-m^2+8=8m+24\ge0\Rightarrow m\ge-3\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+4\right)\\x_1x_2=m^2-8\end{matrix}\right.\)
\(A=x_1^2+x_2^2+2x_1x_2-2x_1x_2-\left(x_1+x_2\right)\)
\(A=\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)\)
\(A=4\left(m+4\right)^2-2\left(m^2-8\right)-2\left(m+4\right)\)
\(A=2m^2+30m+72\)
\(A=2\left(m+3\right)\left(m+12\right)\)
Do \(m\ge-3\Rightarrow\left\{{}\begin{matrix}m+3\ge0\\m+12>0\end{matrix}\right.\) \(\Rightarrow2\left(m+3\right)\left(m+12\right)\ge0\)
\(\Rightarrow A_{min}=0\) khi \(m=-3\)
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