Giải thích các bước giải:
$a)2x^2-8x-11$
$=2(x^2-4x-\dfrac{11}{2})$
$=2(x^2-4x+4-4-\dfrac{11}{2})$
$=2[(x-2)^2-\dfrac{19}{2})$
$=2(x-2)^2-19$
$\text{Ta có:}$
$2(x-2)^2≥0$ $∀x∈R$
$⇒2(x-2)^2-19≥-19$ $∀x∈R$
$\text{Dấu "=" xảy ra khi:}$
$2(x-2)^2-19=-19$
$⇔2(x-2)^2=0$
$⇔(x-2)^2=0$
$⇔x-2=0$
$⇔x=2$
$\text{Vậy $MIN_{(A)}=-19$ tại $x=2$}$
$b)9-15x-3x^2$
$=-3(x^2+5x-3)$
$=-3[x^2+5x+(\dfrac{5}{2})^2-(\dfrac{5}{2})^2-3]$
$=-3[(x+\dfrac{5}{2})^2-\dfrac{37}{4}]$
$=-3(x+\dfrac{5}{2})^2+\dfrac{111}{4}$
$\text{Ta có:}$
$-3(x+\dfrac{5}{2})^2≤0$ $∀x∈R$
$⇒-3(x+\dfrac{5}{2})^2+\dfrac{111}{4}≤\dfrac{111}{4}$ $∀x∈R$
$\text{Dấu "=" xảy ra khi:}$
$-3(x+\dfrac{5}{2})^2+\dfrac{111}{4}=\dfrac{111}{4}$
$⇔-3(x+\dfrac{5}{2})^2=0$
$⇔(x+\dfrac{5}{2})^2=0$
$⇔x+\dfrac{5}{2}=0$
$⇔x=-\dfrac{5}{2}$
$\text{Vậy $MAX_{(B)}=\dfrac{111}{4}$ tại $x=-\dfrac{5}{2}$}$
Học tốt!!!
