Tìm GTNN(hoặc GTLN)của biểu thức sau:a,A=2x^2-8x-11. b,B=9-15x-3x^2

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!!!