toán đại số lớp 7

N

nhuquynhdat

$A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+...+\dfrac{1}{105^2}$

$A< \dfrac{1}{100.101}+\dfrac{1}{101.102}+...+\dfrac{1}{104.105}$

$A< \dfrac{1}{100}-\dfrac{1}{105}=\dfrac{5}{105 00}=\dfrac{1}{2100}=\dfrac{1}{2^2.3.5^2.7}$

\Rightarrow $A<B$
 
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