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a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)
\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)
\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)
\(C=\frac{3.8.15....80.99}{4.9.16.81.100}\)
\(=\frac{1.3.2.4.3.5...8.10.9.11}{2.2.3.3.4.4...9.9.10.10}\)
\(=\frac{\left(1.2.3....9\right).\left(3.4.5...10.11\right)}{\left(2.3.4.5...10\right).\left(2.3.4...10\right)}\)
\(=\frac{11}{10}\)
trả lời
c=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{99}{100}\)
C=\(\frac{3.8.15....99}{4.9.16.100}\)
C=\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\)
C=\(\frac{\left(1.2.....9\right)}{2.3....10}.\left(\frac{3.4....11}{2.3...10}\right)\)
C=\(\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
\(\frac{-15}{14}.\frac{-28}{45}\)
\(=\frac{15.28}{14.15}=\frac{14.15.2}{14.15}=2\)
\(\frac{-15}{14}\).\(\frac{-28}{45}\)=\(\frac{-15.\left(-28\right)}{14.15}\)=\(\frac{-1.\left(-2\right)}{1.3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}\)
=\(\frac{1.3.2.4.3.5....999.101}{2.2.3.3.4.4....100.100}=\frac{1.101}{2.100}=\frac{101}{200}\)
\(=\frac{3+\frac{3}{4}-\frac{3}{19}}{\frac{5}{2}+\frac{5}{8}-\frac{5}{38}}.\frac{15+\frac{15}{34}+\frac{15}{17}-\frac{15}{144}}{2+\frac{2}{34}+\frac{2}{17}-\frac{2}{144}}+2018\)
\(=\frac{3\left(1+\frac{1}{4}-\frac{1}{19}\right)}{\frac{5}{2}\left(1+\frac{1}{4}-\frac{1}{19}\right)}.\frac{15\left(1+\frac{1}{34}+\frac{1}{17}-\frac{1}{144}\right)}{2\left(1+\frac{1}{34}+\frac{1}{17}-\frac{1}{144}\right)}+2018\)
\(=\frac{3}{\frac{5}{2}}.\frac{15}{2}+2018\)
\(=\frac{6}{5}.\frac{15}{2}+2018\)
\(=9+2018=2027\)
\(\frac{3+0,75-\frac{3}{19}}{\frac{5}{2}+0,625-\frac{5}{38}}\) . \(\frac{15+\frac{15}{17}-\frac{15}{144}+\frac{15}{34}}{2+\frac{2}{17}-\frac{2}{144}+\frac{1}{17}}\)+2018
=\(\frac{3+\frac{3}{4}-\frac{3}{19}}{\frac{5}{2}+\frac{5}{8}-\frac{5}{38}}\) . \(\frac{15+\frac{15}{17}-\frac{15}{144}+\frac{15}{34}}{2+\frac{2}{17}-\frac{2}{144}+\frac{2}{34}}\)+2018
=\(\frac{3\left(1+\frac{1}{4}-\frac{1}{19}\right)}{5\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}\) . \(\frac{15\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}{2\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}\)+2018
=\(\frac{6\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}{5\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}\) . \(\frac{15\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}{2\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}\)+2018
=\(\frac{6}{5}\) . \(\frac{15}{2}\)+2018
=9+2018
=2027
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