Giải thích các bước giải:
$P=-x^2+3x+1$
$=-(x^2-3x-1)$
$=-\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2+1\right]$
$=-\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}\right)$
$=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}$
Ta có:
$-\left(x-\dfrac{3}{2}\right)^2≤0$ $∀x$
$⇒-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}≤\dfrac{5}{4}$ $∀x$
Dấu '=' xảy ra khi:
$x-\dfrac{3}{2}=0$
$⇒x=\dfrac{3}{2}$
Vậy $P_{(min)}=\dfrac{5}{4}$ tại $x=\dfrac{3}{2}$
