[toán 9] tìm min

H

huynhbachkhoa23

$a=\sqrt{x} \ge 0$

$A=\dfrac{a^2-a-11}{a+1}=\dfrac{a^2+a-2a-2-9}{a+1}=a-2-\dfrac{9}{a+1} \ge -2-\dfrac{9}{1}=-11$

Đẳng thức xảy ra khi $a=0$
 
E

eye_smile

$A=\dfrac{-11(a+1)+a^2+10a}{a+1}=-11+\dfrac{a(a+10)}{a+} \ge -11$ do $a \ge 0$

Dấu"=" xảy ra \Leftrightarrow $a=0$

\Leftrightarrow $x=0$
 
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