[Toán 9] Tìm min

V

vanmanh2001

Theo bài ra ta có
$\sqrt{(5x-2)^2} + \sqrt{(5x)^2}$
$= |5x-2| + |5x| = |5x-2| + |-5x| \geq |5x- 2 - 5x| = |2| = 2$
Dấu "=" xảy ra khi
$(5x-2)(-5x) \geq 0 $
$\Rightarrow 0 \leq x \leq \frac{2}{5} $
 
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T

transformers123

$A=\sqrt{25x^2-20x+4}+\sqrt{25x^2}$

$\iff A= \sqrt{(5x-2)^2}+\sqrt{(5x)^2}$

$\iff A=|5x-2|+|5x|$

$\iff A=|2-5x|+|5x|$

$\iff A \ge |2-5x+5x|$

$\iff A \ge |2|$

$\iff A \ge 2$

Dấu "=" xảy ra khi $(2-5x).5x \ge 0 \iff 0 \le x \le \dfrac{2}{5}$
 
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