\(A=\left(x_1+x_2\right)^2-3x_1x_2-9\)
\(=3\left(m+2\right)^2-3\left(m-1\right)-9\)
\(=3\left(m^2+4m+4\right)-3m+3-9\)
\(=3m^2+12m+12-3m-6\)
\(=3m^2+9m+6\)
\(=3\left(m^2+3m+2\right)\)
\(=3\left(m^2+3m+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=3\left(m+\dfrac{3}{2}\right)^2-\dfrac{3}{4}>=-\dfrac{3}{4}\forall m\)
Dấu '=' xảy ra khi \(m+\dfrac{3}{2}=0\)
=>\(m=-\dfrac{3}{2}\)
Ta có:
\(A=\left(x_1+x_2\right)^2-3x_1x_2-9\)
\(A=\left(m+2\right)^2-3\left(m-1\right)-9\)
\(A=m^2+4m+4-3m+3-9\)
\(A=m^2+m-2\)
\(A=m^2+2\cdot\dfrac{1}{2}m\cdot1+\dfrac{1}{4}-\dfrac{9}{4}\)
\(A=\left(m+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
Vì \(\left(m+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(m+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(m+\dfrac{1}{2}\right)^2=0\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy min A= \(-\dfrac{9}{4}\Leftrightarrow m=-\dfrac{1}{2}\)
