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\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)
Học good
\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}\)
\(=-\frac{101}{200}\)
\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)
Hok tốt
B = \(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)
B = \(\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)
B = \(-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)
B = \(-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)
B = \(-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)
B = \(-\left(\frac{1}{10}.\frac{11}{2}\right)\)
B = \(\frac{-11}{20}\)
Vì \(\frac{11}{20}>\frac{11}{21}\)nên \(\frac{-11}{20}< \frac{-11}{21}\)
Vậy \(B< \frac{-11}{21}\)
B=\(\frac{9}{100}\)
Dũng, bạn ghi chi tiết cách giải giúp mk nha. Cám ơn bạn nhiều