Question

Tìm GTNN của A = \(x^2+3x+1\)

Answer 1

Ta có :

\(x^2+3x+1=x^2+\dfrac{3}{2}x+\dfrac{3}{2}x+1\\ =x\cdot\left(x+\dfrac{3}{2}\right)+\dfrac{3}{2}\cdot\left(x+\dfrac{3}{2}\right)-\dfrac{5}{4}\\ =\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\)

Ta thấy : \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\forall x\\ \Leftrightarrow A\ge-\dfrac{5}{4}\forall x\\ \Rightarrow Min_A=-\dfrac{5}{4}\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0hayx=-\dfrac{3}{2}\\ VậyMin_A=-\dfrac{5}{4}tạix=-\dfrac{3}{2}\)

P/s : Bn ơi, Toán lp 7 mà bn!