Bài làm
\(\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}{\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)}\)
\(=\frac{\left(\frac{2}{2}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)}{\left(\frac{2}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right)}{\frac{1}{2}\left(2-1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}\right)}\)
\(=\frac{3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}{1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}}\)
\(=\frac{\frac{24}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}{\frac{8}{8}+\frac{4}{8}-\frac{2}{8}+\frac{1}{8}}\)
\(=\frac{31}{8}\div\frac{11}{8}\)
\(=\frac{31}{8}\cdot\frac{8}{11}\)
\(=\frac{31}{11}\)
P/S: Trông không thuận tiện lắm :/
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