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a)\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}=\frac{101}{2.100}=\frac{101}{200}\)
b)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{3599}{3600}=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.6}.....\frac{59.61}{60.60}=\frac{2.61}{60}=\frac{61}{30}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16...10000}=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}=\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(\frac{1.101}{100.2}=\frac{101}{200}\)
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
Ta có: \(\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}\); \(\frac{8}{9}=1-\frac{1}{9}=1-\frac{1}{3^2}\)
\(\frac{15}{16}=1-\frac{1}{16}=1-\frac{1}{4^2}\); ...; \(\frac{9999}{10000}=1-\frac{1}{10000}=1-\frac{1}{100^2}\)
=> \(C=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=99-B\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
=> A > 99-1 = 98
=> B > 98