\(D=ax^2+bx+c\)
\(D=a\left(x^2+\frac{bx}{a}+\frac{c}{a}\right)\)
\(D=a\left(x^2+2\cdot x\cdot\frac{b}{2a}+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\)
\(D=a\left[\left(x+\frac{b}{2a}\right)^2+\frac{4ca-b^2}{4a^2}\right]\)
\(D=a\left(x+\frac{b}{2a}\right)^2+\frac{4ca-b^2}{4a}\ge\frac{4ca-b^2}{4a}\forall x;a>0\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-b}{2a}\)
Ta có \(x^2\ge0\)
\(\Rightarrow ax^2\ge0\left(a>0\right)\)
nên để \(ax^2\)nhỏ nhất thì \(x=0\)
Khi đó \(GTNN_D=a.0^2+b.0+c=c\)
Ta có: \(D=a\left(x^2+\frac{b}{a}x\right)+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)\)
Đặt \(c-\frac{b^2}{4a}=k\). Do \(\left(x+\frac{b^2}{2a}\right)^2\ge0\) nên:
Nếu a > 0 thì \(a\left(x+\frac{b}{2a}\right)^2\ge0\) do đó \(D\ge k\Rightarrow min_D=k\Leftrightarrow x=-\frac{b}{2a}\)
