tính:
\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right).\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
Những câu hỏi liên quan
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
2.Tính:
Aleft(dfrac{1}{10}-1right).left(dfrac{1}{11}-1right)left(dfrac{1}{12}-1right).....left(dfrac{1}{100}-1right)
Bleft(1-dfrac{1}{2^2}right)left(1-dfrac{1}{3^2}right)left(1-dfrac{1}{10^2}right)
Cleft(dfrac{7}{9}+1right)left(dfrac{7}{20}+1right)left(dfrac{7}{33}+1right)...left(dfrac{7}{108080}+1right)
Dleft(1-dfrac{28}{10}right)left(1-dfrac{52}{22}right)left(1-dfrac{80}{36}right)...left(1-dfrac{21808}{10900}right)
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2.Tính:
A=\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right).....\left(\dfrac{1}{100}-1\right)\)
B=\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{10^2}\right)\)
C=\(\left(\dfrac{7}{9}+1\right)\left(\dfrac{7}{20}+1\right)\left(\dfrac{7}{33}+1\right)...\left(\dfrac{7}{108080}+1\right)\)
D=\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
22 tháng 1 2022 lúc 15:10
Tính A=\(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1+\dfrac{1}{100}\right)\)
\(A=\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot...\cdot\dfrac{1}{2}=\dfrac{1}{100}\)
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Tính một cách hợp lý:
aleft(dfrac{1}{2}-1right)left(dfrac{1}{3}-1right)...left(dfrac{1}{100}-1right)) x:dfrac{99}{100}:dfrac{98}{99}:...:dfrac{2}{3}:dfrac{1}{2}
b) dfrac{5-dfrac{5}{3}+dfrac{5}{9}-dfrac{5}{27}}{8-dfrac{8}{3}+dfrac{8}{9}-dfrac{8}{27}}:dfrac{15-dfrac{15}{11}+dfrac{15}{121}}{16-dfrac{16}{11}+dfrac{16}{121}}
c) dfrac{dfrac{1}{9}-dfrac{5}{6}-4}{dfrac{7}{12}-dfrac{1}{36}-10}
d) left(dfrac{1}{2}+1right)left(dfrac{1}{3}+1right)...left(dfrac{1}{99}+1right)
e)
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Tính một cách hợp lý:
a\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)\)) \(x:\dfrac{99}{100}:\dfrac{98}{99}:...:\dfrac{2}{3}:\dfrac{1}{2}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
c) \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
d) \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{99}+1\right)\)
e)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
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25 tháng 1 2022 lúc 20:57
Tính hợp lí:
a) (100-9)(99-9)(98-9)...(1-9).
b) \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
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12 tháng 5 2022 lúc 20:43
\(\dfrac{99}{100}:\left(\dfrac{1}{4}-\dfrac{1}{12}+\dfrac{1}{3}\right)-\left(\dfrac{-7}{5}\right)^2\)
\(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
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A=\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right).......\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)
Lời giải:\(A=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
\(\Rightarrow A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right)}{2.3.4....99.100}\)
\(\Rightarrow A=\dfrac{1}{100}\)
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\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)....\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
So sánh B với \(\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
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\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
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tính:\(C=\left(1-\dfrac{1}{2^2}\right)\left(\dfrac{1}{3^2}-1\right)\left(1-\dfrac{1}{4^2}\right)\left(\dfrac{1}{5^2}-1\right)...\left(\dfrac{1}{99^2}-1\right)\left(1-\dfrac{1}{100^2}\right)\)
\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)
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