Đáp án : Bài 1: $B>A$
Giải thích các bước giải:
Bài 1:
Ta có :
$A=\dfrac{2}{60\times 63}+\dfrac{2}{63\times 66}+...+\dfrac{2}{117\times 120}+\dfrac{2}{120\times 123}$
$\rightarrow A=2(\dfrac{1}{60\times 63}+\dfrac{1}{63\times 66}+...+\dfrac{1}{117\times 120}+\dfrac{1}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{3}{60\times 63}+\dfrac{3}{63\times 66}+...+\dfrac{3}{117\times 120}+\dfrac{3}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{63-60}{60\times 63}+\dfrac{66-63}{63\times 66}+...+\dfrac{120-117}{117\times 120}+\dfrac{123-120}{120\times 123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+..+\dfrac{1}{117}-\dfrac{120}+\dfrac{1}{120}-\dfrac{1}{123})$
$\rightarrow A=\dfrac{2}{3}(\dfrac{1}{60}-\dfrac{1}{123})$
Lại có :
$B=\dfrac{5}{40\times 44}+\dfrac{5}{44\times 48}+..+\dfrac{5}{76\times 80}+\dfrac{5}{2006}$
$\rightarrow B=5(\dfrac{1}{40\times 44}+\dfrac{1}{44\times 48}+..+\dfrac{1}{76\times 80}+\dfrac{1}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{4}{40\times 44}+\dfrac{4}{44\times 48}+..+\dfrac{4}{76\times 80}+\dfrac{4}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{44-40}{40\times 44}+\dfrac{48-44}{44\times 48}+..+\dfrac{80-76}{76\times 80}+\dfrac{4}{2006})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+..+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{2}{80}-\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B=\dfrac{5}{4}.(\dfrac{1}{80}+\dfrac{2}{1003})$
$\rightarrow B>1>\dfrac{1}{2}.\dfrac{1}{60}>A$
