Tìm GTNN của biểu thức

A=4x^2+2x+3

\(A=4x^2+2x+3\)

\(=\left(4x^2+2x+\dfrac{1}{4}\right)+\dfrac{11}{4}\)

\(=\left(2x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Do : \(\left(2x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(2x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Vậy \(GTNN\) của \(A=\dfrac{11}{4}\) . Dấu \("="\) xảy ra khi : \(\left(2x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=-\dfrac{1}{4}\)

\(A=2x^2+4x=2x^2+4x+2-2=2\left(x^2+2x+1\right)-2=2\left(x+1\right)^2-2,tacó2\left(x+1\right)^2\ge0\Leftrightarrow2\left(x+1\right)^2-2\ge-2.VậyminA=-2khix+1=0\Rightarrow x=-1\)\(B=2x^2+4x+3=\left(2x^2+4x+2\right)+1=2\left(x^2+2x+1\right)+1=2\left(x+1\right)^2+1,vì2\left(x+1\right)^2\ge0\Leftrightarrow2\left(x+1\right)^2+1\ge1.VậyminB=1khix=-1\)