\(\text{a) }A=2x^2+4x\)
\(A=2x^2+4x+2-2\)
\(A=2\left(x^2+2x+1\right)-2\)
\(A=2\left(x+1\right)^2-2\)
\(\text{Vì }2\left(x+1\right)^2\ge0\)
\(\text{nên }2\left(x+1\right)^2-2\ge-2\)
\(\text{hay }A\ge0\)
\(\text{Vậy }GTNN_A=-2\text{, dấu bằng xảy ra khi x = -1}\)
\(A=2x^2+4x=2\left(x^2+2x\right)\)
\(=2\left(x^2+2x+1-1\right)\)
\(=2\left[\left(x+1\right)^2-1\right]\)
\(=2\left(x+1\right)^2-2\ge-2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-1\)
\(B=-9x^2+6x-6\)
\(B=-\left(9x^2-6x+6\right)\)
\(B=-\left[\left(3x\right)^2-2\cdot3x\cdot1+1+5\right]\)
\(B=-\left[\left(3x-1\right)^2+5\right]\)
\(B=-5-\left(3x-1\right)^2\le-5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{3}\)
\(B=-9x^2+6x-6\)
\(B=-9x^2+6x-1-5\)
\(B=-\left(9x^2-6x+1\right)-5\)
\(B=-\left(3x-1\right)^2-5\)
\(\text{Vì }-\left(3x-1\right)^2\le0\)
\(\text{nên }-\left(3x-1\right)^2-5\le-5\)
\(\text{hay }B\le-5\)
\(\text{Vậy }GTLN_B=-5\text{, dấu bằng xảy ra khi x = }\frac{1}{3}\)
