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thaotran19

Đặt $A$=$\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{8}$+ $\dfrac{1}{16}$+$\dfrac{1}{32}$+$\dfrac{1}{64}$
\Rightarrow $2A$=$1$+$\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{8}$+ $\dfrac{1}{16}$+$\dfrac{1}{32}$
\Rightarrow $A$=$2A-A=1$-$\dfrac{1}{64}$=$\dfrac{63}{64}$
:)>-:)>-Chúc em năm mới thêm 1 tuổi ,học giỏi thêm nhé!!
 
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thannonggirl

Bài giải

Gọi $A$=$\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{8}$+ $\dfrac{1}{16}$+$\dfrac{1}{32}$+$\dfrac{1}{64}$
\Rightarrow $2A$=2($\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{8}$+ $\dfrac{1}{16}$+$\dfrac{1}{32}$+$\dfrac{1}{64}$)
\Rightarrow $A$$1$+$\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{1}{8}$+ $\dfrac{1}{16}$+$\dfrac{1}{32}$
\Rightarrow $A$=$2A-A=1$-$\dfrac{1}{64}$
\Rightarrow $A$=$\dfrac{63}{64}$
 
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