Đáp án:
Giải thích các bước giải:
$1.6x-x^2-5$
$=-x^2+6x-5$
$=-x^2+6x-9+4$
$=-(x-3)^2+4$
$-(x-3)^2 \leq 0∀x ⇒ -(x-3)^2+4 \leq 4∀x$
$⇒Max=4 ⇔ x-3=0 ⇔ x=3$
$2.4x-x^2+3$
$=-x^2+4x+3$
$=-x^2+4x-4+7$
$=-(x^2-4x+4)+7$
$=-(x-2)^2+7$
$-(x-2)^2 \leq 0∀x ⇒ -(x-2)^2+7 \leq 7∀x$
$⇒Max=7 ⇔ x-2=0 ⇔ x=2$
$3.x-x^2$
$=-x^2+x-\dfrac{1}{4}+\dfrac{3}{4}$
$=-\left (x^2-x+\dfrac{1}{4} \right )+\dfrac{3}{4}$
$=-\left (x-\dfrac{1}{2} \right )^2+\dfrac{3}{4}$
$-\left (x-\dfrac{1}{2} \right )^2 \leq 0∀x ⇒ -\left (x-\dfrac{1}{2} \right )^2+\dfrac{3}{4} \leq \dfrac{3}{4}∀x$
$⇒Max=\dfrac{3}{4} ⇔ x-\dfrac{1}{2}=0 ⇔ x=\dfrac{1}{2}$
Chúc em học tốt.
