Đáp án:
$M_{\min}=-\dfrac{1}{4}$
Giải thích các bước giải:
$M=x^2+9x+20$
$=x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}-\dfrac{1}{4}$
$=\left(x+\dfrac{9}{2}\right)^2-\dfrac{1}{4}$
Ta có:
$\left(x+\dfrac{9}{2}\right)^2\ge 0$
$⇒\left(x+\dfrac{9}{2}\right)^2-\dfrac{1}{4}\ge -\dfrac{1}{4}$
$⇒M\ge -\dfrac{1}{4}⇒M_{\min}=-\dfrac{1}{4}$
Dấu "=" xảy ra khi:
$\left(x+\dfrac{9}{2}\right)^2= 0$
$⇒x=-\dfrac{9}{2}$
Vậy $M_{\min}=-\dfrac{1}{4}$ khi $x=-\dfrac{9}{2}$.
