\(x^2-x\left(m+2\right)-m-3=0\)
Ta có \(\Delta=\left(m+2\right)^2-4\cdot\left(-m-3\right)\ge0\)
\(\Leftrightarrow m^2+8m+16\ge0\)
\(\Leftrightarrow\left(m+4\right)^2\ge0\) ( luôn đúng )
Theo định lý Vi-ét ta có:
\(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=-m-3\end{matrix}\right.\)
Khi đó \(A=-x_1^2x_2-x_1x_2^2=-x_1x_2\left(x_1+x_2\right)=\left(m+2\right)\left(m+3\right)\)
\(A=m^2+5m+6\)
\(A=\left(m+\frac{5}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall m\)
Vậy \(A_{min}=\frac{-1}{4}\Leftrightarrow m=\frac{-5}{2}\)
