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Đáp án:
$min_{D}=\dfrac{7}{4}$ khi $x=-\dfrac{5}{2}$.
Giải thích các bước giải:
$D=x(x+5)+8$
$=x^2+5x+8$
$=x^2+5x+\dfrac{25}{4}+\dfrac{7}{4}$
$=x^2+\dfrac{5}{2}x+\dfrac{5}{2}x+\dfrac{25}{4}+\dfrac{7}{4}$
$=x\bigg{(}x+\dfrac{5}{2}\bigg{)}+\dfrac{5}{2}.\bigg{(}x+\dfrac{5}{2}\bigg{)}+\dfrac{7}{4}$
$=\bigg{(}x+\dfrac{5}{2}\bigg{)}.\bigg{(}x+\dfrac{5}{2}\bigg{)}+\dfrac{7}{4}$
$=\bigg{(}x+\dfrac{5}{2}\bigg{)}^2+\dfrac{7}{4}$
Vì $\bigg{(}x+\dfrac{5}{2}\bigg{)}^2 \geq 0 ⇒ \bigg{(}x+\dfrac{5}{2}\bigg{)}^2+\dfrac{7}{4} \geq \dfrac{7}{4} ⇒ D\geq \dfrac{7}{4}$
Dấu "=" xảy ra khi: $\bigg{(}x+\dfrac{5}{2}\bigg{)}^2+\dfrac{7}{4}=\dfrac{7}{4}$
$⇒\bigg{(}x+\dfrac{5}{2}\bigg{)}^2=0 ⇒ x+\dfrac{5}{2}=0 ⇒ x=-\dfrac{5}{2}$
Vậy $min_{D}=\dfrac{7}{4}$ khi $x=-\dfrac{5}{2}$.
