toán 9] Tìm giá trị nhỏ nhất

V

vipboycodon

A = $\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}$
= $\sqrt{(\sqrt{x-4}+2)^2}+\sqrt{(\sqrt{x-4}-2)^2}$
= $|\sqrt{x-4}+2|+|\sqrt{x-4}-2|$
= $|\sqrt{x-4}+2|+|2-\sqrt{x-4}|$
$\ge |\sqrt{x-4}+2+2-\sqrt{x-4}| = 4$
Dấu "=" xảy ra khi $4 \le x \le 8$
 
D

duchieu300699

$A=\sqrt[]{x-4+4\sqrt[]{x-4}+4}+\sqrt[]{x-4-4\sqrt[]{x-4}+4}$
\Leftrightarrow $A=\sqrt[]{x-4}+2+|\sqrt[]{x-4}-2|$ \geq $|2-\sqrt[]{x-4}+\sqrt[]{x-4}+2|=4$
Đẳng thức xảy ra khi $\sqrt[]{x-4}$\leq $2$
\Leftrightarrow $4$\leq$x$\leq $8$
:)>-
 
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