[Toán 9] GTLN

L

lp_qt

$x \ge 9$

$\sqrt{x-9}=\dfrac{1}{3}.\sqrt{9(x-9)} \le \dfrac{1}{3}.\dfrac{x-9+9}{2}=\dfrac{1}{6}.x$

$\rightarrow \dfrac{\sqrt{x-9}}{5x} \le \dfrac{\dfrac{1}{6}x}{5x}=\dfrac{1}{30}$ khi $x=18$
 
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