Tính \(A\) :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
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Tính:\(\frac{1}{x}+\frac{1}{x+1}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
Thực hiện phép tính :
\(\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
khó quá làm sao mà trả lời đc
tự đầu mình vắt óc mà suy nghĩ
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Tính nhanh:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
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\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)
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\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
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Tính nhanh : { 2 cách } A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
Ta có: \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
=>2A=\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\)
=>2A-A=(\(1+\frac{1}{2^2}+...+\frac{1}{2^4}+\frac{1}{2^5}\))--(\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\))
=>A=\(1-\frac{1}{2^6}\)
=>A=\(\frac{63}{64}\)
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A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Nhận xét :
1/2 = 1 - 1/2 ; 1/4 = 1/2 - 1/4 ; 1/8 = 1/4 - 1/8 ; ..... ; 1/256 = 1/128 - 1/256
=> A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/128 - 1/256
=> A = 1 - 1/256 = 255/256
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\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{156}\)
Ta có:
\(\frac{1}{2}=1-\frac{1}{2}\) \(;\) \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(;\) \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)\(;...;\) \(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{256}{256}-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Vậy \(A=\frac{255}{256}\)
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\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
quy đồngcác phân số lấy mẫu số là 512 .ta có tử số là
256 +128 + 64 +32 +16 +8 +4 +2 +1 =495
A =\(\frac{495}{512}\)
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cho hỏi làm thế nào để nó ra phân số như thế kia zạ
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Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiềua)frac{1+frac{1}{2}+frac{1}{3}+frac{1}{4}}{1-frac{1}{2}+frac{1}{3}-frac{1}{4}}:frac{3+frac{3}{2}+frac{3}{3}+frac{3}{4}}{2-frac{2}{2}+frac{2}{3}-frac{2}{4}}+frac{1}{3}b) frac{frac{1}{3}-frac{1}{5}-frac{1}{7}}{frac{2}{3}-0,4-frac{2}{7}}+frac{frac{3}{8}-frac{3}{16}-frac{3}{32}+frac{3}{64}}{frac{1}{4}-frac{1}{8}-frac{1}{16}+frac{1}{32}}c) frac{0,4-frac{2}{9}+frac{2}{11}}{1,4-frac{7}{9}+frac{7}{11}}-frac{frac{1}{3}-0,25+frac{1}{5}}{1frac{1}{6}-0...
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Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều
a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)
b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)
c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
Tính:\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...\)
\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\)
=>\(\frac{S}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}}\)
=> \(\frac{S}{2}-S=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{n+1}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^n}\right)\)
=> \(-\frac{S}{2}=\frac{1}{2^{n+1}}-1\)
=> S= \(2-\frac{1}{2^n}\)
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