\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
có tử = 1 thì k bt à nha
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{32}=\frac{31}{32}\)
mà \(A=\frac{1}{x}\)
nên \(\frac{1}{x}=\frac{31}{32}\)
\(\Rightarrow\)\(x=\frac{32}{31}\)
Vậy...
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}.\)
\(=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)+\left(\frac{1}{16}+\frac{1}{32}\right)\)
\(=\frac{7}{8}+\frac{3}{32}\)
\(=\frac{31}{32}\)
\(\Leftrightarrow x\)không tồn tại
Đặt \(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(\Rightarrow2S-S=1-\frac{1}{2^5}\)
\(S=1-\frac{1}{32}=\frac{31}{32}=\frac{1}{x}\)
\(\Rightarrow x=\frac{32}{31}\)
