\(A=3x^2+2x-5\)

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

\(=\left[\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\dfrac{1}{\sqrt{3}}+\left(\dfrac{1}{\sqrt{3}}\right)^2\right]-\dfrac{1}{3}-5\)

\(=\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{16}{3}\)

Ta có :

\(\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\) với mọi x

Dấu = xảy ra khi

\(\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2=0\)

\(\Leftrightarrow\sqrt{3}x+\dfrac{1}{\sqrt{3}}=0\)

\(\Leftrightarrow\sqrt{3}x=-\dfrac{1}{\sqrt{3}}\Leftrightarrow-\dfrac{1}{3}\)

Vậy \(Min_A=-\dfrac{16}{3}\Leftrightarrow x=-\dfrac{1}{3}\)