\(\frac{3}{7×10}+\frac{3}{10×13}+......+\frac{3}{97×1001}\)
Cái cuối là 97×100 nha, cậu nào giải hộ mk cài ạ
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\(\left(1-\frac{3}{4}\right).\left(1-\frac{3}{7}\right).\left(1-\frac{3}{10}\right).\left(1-\frac{3}{13}\right)...\left(1-\frac{3}{97}\right).\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1.4.7.10...94.97}{4.7.10.13...97.100}=\frac{1}{100}.\)
\(\left[\left(\frac{2}{1001}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1008}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
\(=\left[\left(\frac{4}{2002}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{2016}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
\(=\left(\frac{1}{2002}.\frac{1001}{17}+\frac{33}{34}\right):\left(\frac{25}{2016}.\frac{1008}{25}+\frac{1009}{2016}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{1009}{2016}\right)\)
\(=1:\frac{2017}{2016}\)
\(=\frac{2016}{2017}\)
=10( (1-√4)/(1-4) + (√4-√7)/(4-7)+.....+(√97-√100)/(97-100) )
=10 (1-100)/3
=-990/3 = -330
Mik cx l9
k hay ko tùy bn
\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}=\frac{1}{7}-\frac{1}{100}=\frac{93}{700}\)
\(\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}=\frac{1}{7}-\frac{1}{100}=\frac{93}{700}\)