Tính nhanh:\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4}}}}\)
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\(=\frac{1}{2}+-\frac{1}{3}+\frac{1}{4}+\frac{1}{-5}+\frac{1}{6}+-\frac{1}{2}+\frac{1}{3}+\frac{1}{-4}+\frac{1}{5}\)
\(=\left(\frac{1}{2}+-\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{-4}\right)+\left(\frac{1}{-5}+\frac{1}{5}\right)+\frac{1}{6}\)
\(=0+0+0+0+\frac{1}{6}\)
\(=\frac{1}{6}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{-5}+\frac{1}{6}+\frac{-1}{2}+\frac{1}{3}+\frac{1}{-4}+\frac{1}{5}\)
\(=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{-1}{5}+\frac{1}{6}+\frac{-1}{2}+\frac{1}{3}+\frac{-1}{4}+\frac{1}{5}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{6}\)
\(=0+0+0+0+\frac{1}{6}\)
\(=\frac{1}{6}\)

Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!
\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)
\(=\frac{1}{2}\div4=\frac{1}{8}\)

\(\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{5}\times\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{6}\)
\(=\frac{3+6-2}{12}=\frac{7}{12}\)
\(\frac{1}{2}\)* \(\frac{1}{2}\)+ \(\frac{1}{2}\)*\(\frac{1}{3}\)+ \(\frac{1}{3}\)* \(\frac{1}{4}\)+ \(\frac{1}{4}\)* \(\frac{1}{5}\)+ \(\frac{1}{5}\)* \(\frac{1}{6}\)
=\(\frac{1}{2}\)* \(\frac{1}{6}\)= \(\frac{1}{12}\)
( Những phân số khác nhau bạn loại đi nhé tại mình ko làm được bước đó trên này bạn thông cảm nhé ! )

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1+1+1+1+1+1}{2}\)
\(=\frac{7}{2}\)
Đặt \(T=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{32}+\frac{1}{64}\)
\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)
Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}\)
= 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64
= 63/64

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{481}{280}\)
\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4}}}}\)
\(=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{5}{4}}}}\)
\(=1+\frac{1}{1+\frac{1}{1+\frac{4}{5}}}\)
\(=1+\frac{1}{1+\frac{1}{\frac{9}{5}}}\)
\(=1+\frac{1}{1+\frac{5}{9}}\)
\(=1+\frac{1}{\frac{14}{9}}\)
\(=1+\frac{9}{14}\)
\(=\frac{23}{14}\)