Ta có: \(E=4x^2+4x-5\)
\(=4x^2+4x+1-6\)
\(=\left(2x+1\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
\(A=4x^2+4x-5=4x^2+4x+1-6=\left(2x+1\right)^2-6\)
Do \(\left(2x+1\right)^2\ge0\) \(\Rightarrow\left(2x+1\right)^2-6\ge-6\)
\(\Rightarrow Max\) A=-6\(\Leftrightarrow x=\dfrac{-1}{2}\)