\(A=4x^2+4x+9\)
\(A=4\left(x^2+x+\dfrac{9}{4}\right)\)
\(A=4\left(x^2+2\cdot x\cdot0,5+0,25+2\right)\)
\(A=4\left(x+0,5\right)^2+8\)
Vì \(4\left(x+0,5\right)^2\ge0\forall x\)
\(\Rightarrow4\left(x+0,5\right)^2+8\ge8\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-0,5\)
Vậy \(MIN_A=8\Leftrightarrow x=-0,5\)
\(A=4x^2+4x+9\)
\(A=4x^2+4x+1+8\)
\(A=\left(2x+1\right)^2+8\)
Vì (2x+1)2 + 8 \(\ge\) 0 \(\forall x\) => minA= 8
Dấu "x" xảy ra <=> 2x + 1 = 0
2x = 0 -1
2x = -1
x = \(\dfrac{-1}{2}\)
Vậy minA = 8 khi x = \(-\dfrac{1}{2}\)
// cậu tham khảo //
