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\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}\)
=\(\frac{1.3.2.4.3.5....999.101}{2.2.3.3.4.4....100.100}=\frac{1.101}{2.100}=\frac{101}{200}\)
a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}....\frac{100}{101}\)
Nhận xét: Nếu \(\frac{a}{b}
\(C=\frac{3.8.15....80.99}{4.9.16.81.100}\)
\(=\frac{1.3.2.4.3.5...8.10.9.11}{2.2.3.3.4.4...9.9.10.10}\)
\(=\frac{\left(1.2.3....9\right).\left(3.4.5...10.11\right)}{\left(2.3.4.5...10\right).\left(2.3.4...10\right)}\)
\(=\frac{11}{10}\)
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=\(\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)