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Câu a)
\(B=\frac{\left(5-\frac{5}{13}+\frac{5}{19}-\frac{5}{27}\right)\cdot4024\cdot4026}{\left(10-\frac{10}{13}+\frac{10}{19}-\frac{10}{27}\right)\cdot2012\cdot2013}\)
\(\Leftrightarrow B=\frac{\left(5\cdot1-5\cdot\frac{1}{13}+5\cdot\frac{1}{19}-5\cdot\frac{1}{27}\right)\cdot2012\cdot2\cdot2013\cdot2}{\left(10\cdot1-10\cdot\frac{1}{13}+10\cdot\frac{1}{19}-10\cdot\frac{1}{27}\right)\cdot2012\cdot2013}\)
\(\Leftrightarrow B=\frac{5\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)\cdot2\cdot2}{5\cdot2\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)}\)
\(\Leftrightarrow B=2\)
\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)
\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)
\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)
