\(A=\left(2x-1\right)^2+\left(x+2\right)^2=4x^2-4x+1+x^2+4x+4=5x^2-5x+5\)
\(=5\left(x^2-x+1\right)=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\)
Dấu = xảy ra khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
vậy \(A_{min}=\dfrac{15}{4}khi\) x=1/2