Đáp án:
$-1\le \dfrac{7-6x}{x^2+2} \le \dfrac{9}{2}$
Giải thích các bước giải:
$P=\dfrac{7-6x}{x^2+2}$
$\rightarrow P+1=\dfrac{7-6x}{x^2+2}+1$
$\rightarrow P+1=\dfrac{x^2-6x+9}{x^2+2}$
$\rightarrow P+1=\dfrac{(x-3)^2}{x^2+2}\ge 0\quad \forall x$
$\rightarrow P\ge -1$
$P-\dfrac{9}{2}=\dfrac{7-6x}{x^2+2}-\dfrac{9}{2}$
$\rightarrow P-\dfrac{9}{2}=\dfrac{2(7-6x)-9(x^2+2)}{2(x^2+2)}$
$\rightarrow P-\dfrac{9}{2}=\dfrac{-12x-9x^2-4}{2(x^2+2)}$
$\rightarrow P-\dfrac{9}{2}=\dfrac{-(3x+2)^2}{2(x^2+2)}\le 0\quad \forall x$
$\rightarrow P\le \dfrac{9}{2}$
