[Toán 9] So sánh

H

hoangtrongminhduc

$A=2\left ( \sqrt{1} +\sqrt{3}...+\sqrt{19}\right )$
$B=2\left ( \sqrt{2}+\sqrt{4} ...+\sqrt{20}\right )$
Ta thấy $\sqrt{1}<\sqrt{2}$tương tự $\sqrt{3}<\sqrt{4}$,.... ,$\sqrt{19}<\sqrt{20}$
Cộng vế theo vế ta đc $\left ( \sqrt{1} +\sqrt{3}...+\sqrt{19}\right )<\left ( \sqrt{2}+\sqrt{4} ...+\sqrt{20}\right )$
Suy ra A<B
 
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