[toán 9]so sánh

V

vansang02121998

$A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{17}$$+2\sqrt{19}$

Ta có

$2\sqrt{a} > \sqrt{a-1}+\sqrt{a+1}$

Áp dụng

$A > \sqrt{0}+\sqrt{2}+\sqrt{2}+\sqrt{4}+...+\sqrt{18}+\sqrt{20}$

$A > 2\sqrt{2}+2\sqrt{4}+...+2\sqrt{18}+\sqrt{20}=B$

Vậy, $A>B$
 
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