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\(\frac{6:\frac{3}{5}-1\frac{1}{6}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}=1\)
\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)
\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)
\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)
\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)
\(=\frac{10-1}{\frac{99}{11}}\)
\(=\frac{9}{9}\)
\(=1\)
\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\) \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)
\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\) \(=\) \(\frac{21}{5}\)X \(\frac{10}{11}+\frac{57}{11}\)
\(=\frac{33}{5}-1\) \(=\frac{42}{11}+\frac{57}{11}\)
\(=\frac{28}{5}\) \(=\frac{99}{11}=9\)
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
\(6\frac{3}{5}:6-\frac{1}{8}\times8+\frac{1}{3}\times\frac{3}{50}=\frac{33}{5}\times\frac{1}{6}-1+\frac{1}{50}\)
\(=\frac{33}{30}-1+\frac{1}{50}=\frac{3}{25}\)
Giải: \(6\frac{3}{6}:6-\frac{1}{8}.8+\frac{1}{3}.\frac{3}{50}\)
\(=\frac{13}{2}:6-\frac{1}{8}.8+\frac{1}{3}.\frac{3}{50}\)
\(=\frac{13}{12}-1+\frac{1}{50}\)
\(=\frac{1}{12}+\frac{1}{50}\)
\(=\frac{31}{300}\)
=( 6 x\(\frac{5}{3}\) - \(\frac{7}{6}\) x\(\frac{6}{7}\) ) : (\(\frac{21}{5}\) x\(\frac{10}{11}\) +\(\frac{57}{11}\))
= (10 - 1) :(\(\frac{42}{11}\)+\(\frac{57}{11}\))
=\(\frac{9}{9}\)= 1