(-3)3 - \(\frac{-1}{3}\): \(\frac{4}{15}\)
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\(A=3+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{4}{3}}}}=3+\frac{1}{1+\frac{1}{1+\frac{3}{4}}}\)
\(=3+\frac{1}{1+\frac{1}{\frac{7}{4}}}=3+\frac{1}{1+\frac{4}{7}}=3+\frac{1}{\frac{11}{4}}=3+\frac{4}{11}=\frac{37}{11}\)
\(B=-5+\frac{1}{1-\frac{1}{2+\frac{1}{\frac{3}{4}}}}=-5+\frac{1}{1-\frac{1}{2+\frac{4}{3}}}\)
\(=-5+\frac{1}{1-\frac{1}{\frac{10}{3}}}=-5+\frac{1}{1-\frac{3}{10}}=-5+\frac{1}{\frac{7}{10}}=-5+\frac{10}{7}=\frac{-25}{7}\)
Ta có: \(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(A=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)-\frac{3}{4}-\frac{1}{36}+\frac{1}{73}\)
\(A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{10}{15}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{2}{3}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{6}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{7}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{73}\)
Vậy: \(A=\frac{1}{73}\)
\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)
\(\left[\frac{139}{84}-\frac{577}{924}-\frac{98}{205}\right].\frac{2197}{27}\)
\(\left[\frac{34}{33}-\frac{98}{205}\right]\frac{2197}{27}\)
\(\frac{3736}{6765}.\frac{2197}{27}=44,93713285\)
-103/4
\(=\frac{-103}{4}\)