Question

Bài 1 : Tính

a)              \(1+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}\)

b)                 \(1+1+\frac{1}{1+\frac{3}{1+1}}\)

Answer 1

\(1+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}=1+\frac{1}{1+\frac{1}{\frac{4}{3}}}=1+\frac{1}{1+\frac{3}{4}}=1+\frac{1}{\frac{7}{4}}=1+\frac{4}{7}=\frac{11}{7}\)

\(1+1+\frac{1}{1+\frac{3}{1+1}}=1+1+\frac{1}{1+\frac{3}{2}}=1+1+\frac{1}{\frac{5}{2}}=2+\frac{2}{5}=\frac{12}{5}\)

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Answer 2

a, Ta có \(1+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}=1+\frac{1}{1+\frac{1}{\frac{4}{3}}}=1+\frac{1}{1+\frac{3}{4}}=1+\frac{1}{\frac{7}{4}}=1+\frac{4}{7}=\frac{11}{7}\)

b,Ta có \(1+1+\frac{1}{1+\frac{3}{1+1}}=2+\frac{1}{1+\frac{3}{2}}=2+\frac{1}{\frac{5}{2}}=2+\frac{2}{5}=\frac{12}{5}\)

Answer 3

\(1+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}=1+\frac{1}{1+\frac{1}{\frac{4}{3}}}=1+\frac{1}{1+\frac{3}{4}}=1+\frac{1}{\frac{7}{4}}=1+\frac{4}{7}=\frac{11}{7}\)

\(1+1+\frac{1}{1+\frac{3}{1+1}}=2+\frac{1}{1+\frac{3}{2}}=2+\frac{1}{\frac{5}{2}}=2+\frac{2}{5}=\frac{12}{5}\)