Tính :
A= 256 - \(\dfrac{256}{2}-\dfrac{256}{2^2}-\dfrac{256}{2^3}-....-\dfrac{256}{2^9}\)
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Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)
\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)
\(A=1-\dfrac{1}{256}\)
\(A=\dfrac{255}{256}\)
Lời giải:
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+3}=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^3=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x.(1+\frac{1}{8})=\frac{9}{256}\Rightarrow \left(\frac{1}{2}\right)^x=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow x=5\)
Lời giải:
\(\left(\frac{1}{2}\right)^{\frac{-1}{4}}=(2^{-1})^{\frac{-1}{4}}=2^{\frac{1}{4}}=\sqrt[4]{2}\)
Không đáp án nào đúng.
\(A=256-\dfrac{256}{2}-\dfrac{256}{2^2}-\dfrac{256}{2^3}-.......-\dfrac{256}{2^9}\)
\(\Leftrightarrow A=256\left(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-.....-\dfrac{1}{2^9}\right)\)
Đặt \(B=\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-.....-\dfrac{1}{2^9}\)
\(\Leftrightarrow2B=1-\dfrac{1}{2}-\dfrac{1}{2^2}-.....-\dfrac{1}{2^8}\)
\(\Leftrightarrow2B-B=1-\dfrac{1}{2^9}\)
\(\Leftrightarrow B=1-\dfrac{1}{2^9}\)
\(\Leftrightarrow A=256\left(1-\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow A=256-\dfrac{1}{2^9}\)
\(\Leftrightarrow A=2^8-\dfrac{1}{2^9}\)
\(\Leftrightarrow A=\dfrac{2^{17}}{2^9}-\dfrac{1}{2^9}\)
\(\Leftrightarrow A=\dfrac{2^{17}-1}{2^9}\)
Vậy \(\Leftrightarrow A=\dfrac{2^{17}-1}{2^9}\)
Chúc bạn học tốt >w<