\(F=5+6x+9x^2\)
\(=\left(9x^2+6x+1\right)+4\)
\(=\left(3x+1\right)^2+4\ge4\)
Dấu = xảy ra \(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(Max_F=4\Leftrightarrow x=-\dfrac{1}{3}\)
\(F=5+6x+9x^2\)
\(\Leftrightarrow F=\left(3x\right)^2+2.3x+1+4\)
\(\Leftrightarrow F=\left(3x+1\right)^2+4\)
Do \(\left(3x+1\right)^2\ge0\forall x\)
\(\Rightarrow F=\left(3x+1\right)^2+4\ge4\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow3x+1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy GTNN của P là : \(4\Leftrightarrow x=-\dfrac{1}{3}\)
:D