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\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
=-1+1+-2/11
=0+-2/11
=-2/11
b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)
=1+-1+-5/17
=0+-5/17
=-5/17
c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)
=1+-1+16/17
=0+16/17
=16/17
d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)
=\(\frac{-5}{9}\)+\(\frac{4}{-9}\)+\(\frac{8}{15}\)+\(\frac{7}{15}\)+\(\frac{-2}{11}\)
=(\(\frac{-5}{9}\)+\(\frac{-4}{9}\))+(\(\frac{8}{15}\)+\(\frac{7}{15}\))+\(\frac{-2}{11}\)
=(-1)+1+\(\frac{-2}{11}\)
=0+\(\frac{-2}{11}\)
=\(\frac{-2}{11}\).
1/
a) \(C=\frac{4}{7}.\frac{3}{5}.\frac{7}{4}.\left(-20\right).\frac{5}{6}\)
\(=\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{3}{5}.\frac{5}{6}\right).\left(-20\right)\)
\(=\frac{1}{2}.\left(-20\right)\)
\(=-10\)
2/ \(B=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}.\frac{6^2}{35}.\frac{7^2}{48}.\frac{8^2}{63}.\frac{9^2}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}.\frac{7.7}{6.8}.\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.3.4.5.6.7.8.9}{1.2.3.4.5.6.7.8}.\frac{2.3.4.5.6.7.8.9}{3.4.5.6.7.8.9.10}\)
\(=9.\frac{2}{10}=9.\frac{1}{5}=\frac{9}{5}\)
C=\(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).....\(\frac{9.9}{8.10}\)=\(\frac{2.2.3.3...9.9}{1.3.2.4...8.10}\)= \(\frac{2.9}{1.10}\)= \(\frac{9}{5}\)
\(D.=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(D=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(D=\frac{5}{2}.\frac{98}{99}\)
\(D=\frac{245}{99}\)