Đáp án:
$max_P=\dfrac{7}{3}\Leftrightarrow x=-1$
Giải thích các bước giải:
$P=\dfrac{2x^2+4x+9}{x^2+2x+4}\\ =\dfrac{2x^2+4x+8+1}{x^2+2x+4}\\ =\dfrac{2(x^2+2x+4)+1}{x^2+2x+4}\\ =2+\dfrac{1}{x^2+2x+4}\\ =2+\dfrac{1}{x^2+2x+1+3}\\ =2+\dfrac{1}{(x+1)^2+3}\\ (x+1)^2+3 \ge 3 \ \forall \ x\\ \Rightarrow \dfrac{1}{(x+1)^2+3}\le \dfrac{1}{3} \ \forall \ x\\ \Rightarrow 2+\dfrac{1}{(x+1)^2+3} \le 2+ \dfrac{1}{3} =\dfrac{7}{3}\ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x+1=0 \Leftrightarrow x=-1.$
