Tính(1/4-1).(1/9-1).(1/16-1)....(1/81-1).(1/100-1)
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Làm lại đề cho:
\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)
Tính nhẩm
\(S=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right).\left(\frac{1}{100}-1\right)\)
\(S=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}........\frac{-80}{81}.\frac{-99}{100}\)
\(-S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{80}{81}.\frac{99}{100}\)
\(-S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}........\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(-S=\frac{1.3.2.4.3.5........8.10.9.11}{2.2.3.3.4.4.......9.9.10.10}\)
\(-S=\frac{\left(1.2.3......8.9\right).\left(3.4.5.......10.11\right)}{\left(2.3.4.......9.10\right).\left(2.3.4........9.10\right)}\)\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}=>S=\frac{-11}{20}\)
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\left(9\text{số (-1)}\right)\right].\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right)\right].\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
Ta có :
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow A>\frac{65}{132}\left(đpcm\right)\)
Chúc bạn học tốt !!!!
\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)