Đáp án:
CHÚC BẠN HỌC TỐT!!!
Giải thích các bước giải:
$I=x-\sqrt{x-3}$
$=x-3-2.\sqrt{x-3}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}=0$
$=\bigg{(}\sqrt{x-3}-\dfrac{1}{2}\bigg{)}^2+\dfrac{11}{4}=0$
Vì $\bigg{(}\sqrt{x-3}-\dfrac{1}{2}\bigg{)}^2 \geq0 ⇒ \bigg{(}\sqrt{x-3}-\dfrac{1}{2}\bigg{)}^2+\dfrac{11}{4}\geq\dfrac{11}{4}⇒min_I=\dfrac{11}{4}$
Dấu "=" xảy ra khi: $\sqrt{x-3}=\dfrac{1}{2}⇒x-3=\dfrac{1}{4}⇒x=\dfrac{13}{4}$
$A=(x-1)(4-x)$
$=4x-x^2-4+x$
$=-x^2+5x-4$
$=-\bigg{(}x^2-5x+\dfrac{25}{4}\bigg{)}-\dfrac{9}{4}$
$=-\bigg{(}x-\dfrac{5}{2}\bigg{)}^2-\dfrac{9}{4}$
Vì $-\bigg{(}x-\dfrac{5}{2}\bigg{)}^2\leq0⇒-\bigg{(}x-\dfrac{5}{2}\bigg{)}^2-\dfrac{9}{4}\leq-\dfrac{9}{4}$
$⇒min_A=-\dfrac{9}{4}$
Dấu "=" xảy ra khi: $x-\dfrac{5}{2}=0⇒x=\dfrac{5}{2}$