Ta có: \(x^2+\left(m-1\right)x+m-6=0\)
\(\Delta=m^2-2m+1-4m+24\)
\(\Delta=\left(m-3\right)^2+16>0\forall m\left(\left\{{}\begin{matrix}\left(m-3\right)^2\ge0\forall m\\\Rightarrow\left(m-3\right)^2+16>0\forall m\end{matrix}\right.\right)\)
\(A=\left(x_1x_2\right)^2-4\left[x_1^2+x^2_2\right]+16\)
\(A=\left(x_1x_2\right)^2-4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+16\)
\(A=m^2-12m+52-4\left(m^2-4m+13\right)\)
\(A=-3m^2+4m\)
\(A=-3\left(m^2-2.\frac{2}{3}m+\frac{4}{9}\right)+\frac{4}{3}\)
\(A=-3\left(m-\frac{2}{3}\right)^2+\frac{4}{3}\le\frac{4}{3}\forall m\)
\(\Rightarrow Max_A=\frac{4}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow m=\frac{2}{3}\)
@Băng....
