Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{79}{80}\)và \(\frac{1}{9}\). Hãy so sánh
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A=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{79}{80}.CM:A<\frac{1}{9}\)
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{79}{80}:CM:A<\frac{1}{9}\)
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.....\frac{79}{80}.CMR\)
\(A<\frac{1}{9}\)
Ta đặt B=\(\frac{2}{3}.\frac{4}{5}...\frac{80}{81}\)
Mà \(\frac{1}{2}<\frac{2}{3};\frac{3}{4}<\frac{4}{5};...;\frac{79}{80}<\frac{80}{81}\)
=>A<B
=>A2<AB=\(\frac{1}{2}.\frac{2}{3}.....\frac{80}{81}=\frac{1}{81}\)
=>A2<\(\frac{1}{81}\)
=>A<\(\sqrt{\frac{1}{81}}=\frac{1}{9}\)(đpcm)
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http://olm.vn/hoi-dap/question/419438.html
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Bài 1 : Tính
Cho A =\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+......+\frac{1}{60}>\frac{7}{12}\)
B = \(\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{5^2}+......+\frac{ }{50^{21}}\)
CMR B >\(\frac{1}{4}\)và B < \(\frac{4}{9}\)
C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.......\frac{79}{80}< \frac{1}{9}\)
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}\)
Chứng minh \(A<\frac{1}{9}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}
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Cho: \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}\)
Chứng minh \(A<\frac{1}{9}\)
Cho: \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}\)
Chứng minh \(A<\frac{1}{9}\)
Cho
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}\)
Chứng minh \(A
\(\frac{1.3.5...79}{2.4.6...80}\)= \(\frac{1.3.5...79}{\left(1.2\right).\left(2.2\right).\left(3.2\right)...\left(40.2\right)}\).\(\frac{1.3.5...79}{\left(1.2.3.4...40\right).\left(2.2.2.2...2.2\right)}\)=\(\frac{1.3.5...79}{\left(1.3.5...39\right).\left(2.4.6...40\right).2^{40}}\)<1/9
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Cho \(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{997}{100}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}...\frac{998}{1001}\)
So sánh A và B