\(A=x^2+4x+3=x^2+4x+4-1=\left(x+2\right)^2-1>=-1\forall x\)
Dấu '=' xảy ra khi x+2=0
=>x=-2
\(B=2x^2-5x+1\)
\(=2\left(x^2-\dfrac{5}{2}x+\dfrac{1}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{17}{16}\right)\)
\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{17}{8}>=-\dfrac{17}{8}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{5}{4}=0\)
=>\(x=\dfrac{5}{4}\)