- Đặt \(A=4x^2+4x+5\)
- Ta có: \(A=4x^2+4x+5\)
\(\Leftrightarrow A=\left(4x^2+4x+1\right)+4\)
\(\Leftrightarrow A=\left(2x+1\right)^2+4\)
- Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(2x+1\right)^2+4\ge4\forall x\)
\(\Rightarrow A_{min}=4\)
- Dấu "=" xảy ra khi: \(2x+1=0\)\(\Leftrightarrow\)\(2x=-1\)\(\Leftrightarrow\)\(x=-\frac{1}{2}\left(TM\right)\)
Vậy \(A_{min}=4\)\(\Leftrightarrow\)\(x=-\frac{1}{2}\)